how to solve questions on integration

Evaluate \[\int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx.\] Notice that the integrand is an odd function. So, \[\begin{align*} \int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx &= \int_{-1}^{1} \frac{(-x)^3-2(-x)}{\sqrt{(-x)^4+1}} \, d(-x) \\ &= - \int_{-1}^1 \frac{x^3-2x}{\sqrt{x^4+1}} \, dx \\ &= 0. \end{align*}\] The final equivalence comes from the fact that the integral is equal to the negative of itself. Therefore, it is $0$. $_\square$

\[ \large \int _{ -1 }^{ 1 }{ \frac { \sin { x } }{ 1+{ x }^{ 2 }+{ x }^{ 4 } } \, dx } = \, ? \]

\[\int_a^b f(x) \, dx = \int_a^b f(a+b-x) \, dx.\]

Instead of the function being centered at $0$, the function is now centered at $\tfrac{a+b}{2}$. Then,

\[\int_a^b f(x) \, dx = \frac{1}{2} \int_a^b f(x) + f(a+b-x) \, dx.\]

Evaluate \[ \int_3^7 \frac{\ln(x+2)}{\ln\big(24+10x-x^2\big)} \, dx.\] We have \[\begin{align*} \int_3^7 \frac{\ln(x+2)}{\ln\big(24+10x-x^2\big)} \, dx &= \int_3^7 \frac{\ln(x+2)}{\ln(x+2) + \ln(12-x)} \, dx \\\\ &= \frac{1}{2} \int_3^7 \frac{\ln(12-x) + \ln(x+2)}{\ln(x+2) + \ln(12-x)} \, dx \\\\ &= 2.\ _\square \end{align*}\]

Let $f(x)$ be a real-valued function continuous on $\left[0,2\right]$ such that $f(x)=f(2x)$ for all $x$. If

\[\int_0^1 f(x) dx = 100,\]

then what is the value of

\[\int_0^2 f(x)dx ?\]

Suppose the function $f$ has bounded antiderivative on $[0, \, \infty]$. Then, via the u-substitution $x \mapsto \tfrac{1}{x}$,

\[\int_0^\infty f(x) \, dx = \frac{1}{2} \int_0^\infty f(x) + \frac{f\big(\frac{1}{x}\big)}{x^2} \, dx.\]

Evaluate \[ \int_0^\infty \frac{\ln (2x)}{1 + x^2} \, dx.\] We have \[\begin{align} \int_0^\infty \frac{\ln (2x)}{1 + x^2} \, dx &= \frac{1}{2} \int_0^\infty \frac{\ln(2x) + \ln\big(2x^{-1}\big)}{1 + x^2} \, dx \\\\ &= \frac{2\ln 2}{2} \int_0^\infty \frac{1}{1 + x^2} \, dx \\\\ &= \frac{\pi \ln 2}{2}.\ _\square \end{align}\]

Evaluate the integral $\displaystyle \int_{0}^{\infty} \frac{\ln x}{x^2+2x+4} \, dx.$

Round your answer to three decimal places.

This section is currently incomplete. Let's join hands to build this wiki. Feel free to add anything you know about this topic!

There are more transformations than simply reflections and inversions that maintain the interval of integration, but they are not as common.

Suppose the function $f$ is one-to-one and increasing. Then, a geometric equivalence may be established:

\[\int_a^b f(x) \, dx + \int_{f(a)}^{f(b)} f^{-1}(x) \, dx = bf(b) - af(a).\]

Suppose the function $f$ is one-to-one and decreasing. Then, another geometric equivalence may be established:

\[\int_a^b f(x) \, dx - \int_{f(b)}^{f(a)} f^{-1}(x) \, dx = (b-a)f(b) - a\big(f(a)-f(b)\big).\]

Let $f(x)$ be a one-to-one continuous function such that $f(1)=4$ and $f(6)=2$, and assume $\displaystyle \int_1^6 f(x) \, dx = 15$. Calculate $\displaystyle \int_2^4 f^{-1}(x) \, dx$. The region bounded by $f$, $x = 1$, and $y = 2$ must have area $5$, implying the integral in question corresponds to the area $5 + 1 \cdot (4 - 2) = \boxed{7}$. The above formula for decreasing functions provides the same answer. $_\square$

Integration by parts provides a way to change the integrand directly, and like the exploration of inverse functions, it is a geometric statement. However, this is a statement about the geometry of calculus operators, and any visualization of it would lie in an entirely different space. However, the same intuition can apply. Integration by parts is a very powerful tool, and many problems on this page could be solved by this (and more elementary methods) without the need for anything more complicated.

Integration by parts states that for any differentiable functions $u(x)$ and $v(x)$, the following equivalence holds:

\[ \int u(x) v'(x) \, dx = u(x) v(x) - \int v(x) u'(x) \, dx. \]

This can be thought of as a "backwards" application of the product rule .

Evaluate \[\int_1^7 \ln(1 + x) \, dx.\] Let $u(x) = \ln(1 + x)$ and $v(x) = x + 1$. Then, \[\begin{align} \int_1^7 \ln(1 + x) \, dx &= \big[(x+1)\ln(x+1)\big]_1^7 - \int_1^7 \frac{x+1}{x+1} \, dx \\ &= 8\ln 8 - 2\ln 2 - 6 \\ &= 22 \ln 2 - 6.\ _\square \end{align}\]

The (adjusted) beta function $B(m, \, n)$ is defined for nonnegative integers $m$ and $n$ as \[B(m, \, n) = \int_0^1 x^{m}(1-x)^{n} \, dx.\] Find a closed form for $B(m, \, n)$. Note that $B(0, \, n) = \frac{1}{n+1}$. Now, suppose $n$ is fixed, and note for $m > 0$, \[\begin{align} B(m, \, n) &= \int_0^1 x^m (1-x)^n \, dx \\ &= 0 - \frac{m}{n+1} \int_0^1 x^{m-1} \cdot \left(-(1-x)^{n+1}\right) \, dx \\ &= \frac{m}{n+1} B(m-1, \, n+1). \end{align}\] Thus, $B(m, \, n) = \frac{m}{n+1} \cdot \frac{m-1}{n+2} \cdots \frac{1}{n+m} B(0, n+m) = \frac{m! n!}{(m+n+1)!}.$ $_\square$ Learn more about the beta function (with correctly off-set indices) here .

\[ \int_0^1 \left(1-x^2\right)^9 x^9 \, dx \]

Let $I$ denote the value of the integral above. What is the sum of digits of $I^{-1}?$

When solving integrals with trigonometric functions, trigonometric identities create shortcuts! The Integration Of Trigonometric Functions wiki goes into this in detail, but below are a few examples.

The first identity is $\sin^2x+\cos^2x=1.$

We have \[\begin{align} \int(\sin x+\cos x)^2\, dx &=\int\left(\sin^2x+2\sin x\cos x+\cos^2x\right)\, dx\\ &=\int(1+\sin 2x)\, dx\\ &=x-\dfrac{1}{2}\cos 2x+C, \end{align}\] where $C$ is the constant of integration.

Here is an example of a less obvious application of that identity:

We have \[\begin{align} \int_0^\frac{\pi}{2}\sin^5x\, dx &=\int_0^\frac{\pi}{2}\sin x\left(1-\cos^2x\right)^2\, dx\\ &=-\int_1^0\left(1-u^2\right)^2\, du\\ &=\int_0^1\left(1-u^2\right)^2\, du\\ &=\frac{8}{15}. \end{align}\]

Other examples that can be used are the double-angle formulas , which can be used in the integrals of $\sin^2\theta$ and $\cos^2\theta,$ as well as others.

We have \[\begin{align} \int\frac{\cos 2x}{\sin x+\cos x}\, dx &=\int\frac{\cos^2x-\sin^2x}{\sin x+\cos x}\, dx\\ &=\int(\cos x-\sin x)\, dx\\ &=\sin x+\cos x+C, \end{align}\] where $C$ is the constant of integration.

Finally, the product-to-sum identities help to solve complex integrals.

We have \[\begin{align} \int\sin 2015x \sin 2016x \, dx &=\dfrac{1}{2}\int(\cos x-\cos 4031x)\, dx\\ &=\dfrac{\sin x}{2}-\dfrac{\sin 4031x}{8062}+C, \end{align}\] where $C$ is the constant of integration.

One of the most powerful substitutions using trigonometric functions is the Weierstrass substitution of $t=\tan\frac{\theta}{2}.$ This is most easily seen in rational functions involving trigonometric functions. Through trigonometric identities and manipulation $\sin\theta=\frac{2t}{1+t^2},$ $\cos\theta=\frac{1-t^2}{1+t^2},$ and $d\theta=\frac{2\ dt}{1+t^2}.$ This can often transform the integral into the integral of a rational function, as seen in the following example:

Find the value of $\displaystyle\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}.$ Use the Weierstrass substitution $\cos x=\frac{1-t^2}{1+t^2}$ and $dx=\frac{2dt}{1+t^2}:$ \[\int_0^\frac{\pi}{2}\frac{dx}{2+\cos x}\ =\int_0^1\frac{\frac{2dt}{1+t^2}}{2+\frac{1-t^2}{1+t^2}}=\int_0^1\frac{2}{t^2+3}dt.\] This integral can be finished by a $u$-substitution and application of the derivative of arctangent. $_\square$

The Weierstrass substitution can also be applied to a rather common integral, $\displaystyle\int\csc x\ dx.$ A common method of solving this question is a "clever" multiplication by $\frac{1}{1},$ but the Weierstrass substitution is easier to apply.

We have \[\int\csc \theta\, d\theta=\int\frac{d\theta}{\sin\theta}=\int\frac{\hspace{3mm} \frac{2dt}{1+t^2}\hspace{3mm} }{\frac{2t}{1+t^2}}=\int\frac{dt}{t}=\ln t+C,\] where $C$ is the constant of integration. All that's needed is to re-substitute $t=\tan\frac{\theta}{2}$ to obtain the final value of \[\ln\left(\tan\frac{\theta}{2}\right)+C=-\ln(\cot x+\csc x)+C=\ln\frac{\sin\theta}{\cos\theta+1}+C.\]

A final use of the Weierstrass substitution is the "reverse Weierstrass substitution," which involves simplifying the integral of a rational function with trigonometry.

Find the value of $\displaystyle\int_0^1\dfrac{\arcsin\frac{2x}{1+x^2}}{1+x^2}\, dx$. Recall that in the Weierstrass substitution, $\frac{2x}{1+x^2}=\sin\theta$ and $\frac{2dx}{1+x^2}=d\theta.$ Then the above can be transformed to \[\displaystyle\int_0^\frac{\pi}{2}\frac{\sin^{-1}(\sin\theta)}{2}\, d\theta=\displaystyle\int_0^\frac{\pi}{2}\frac{\theta}{2}\, d\theta.\] This is the integral of a polynomial, which evaluates to $\frac{\pi^2}{16}.$ $_\square$

\[\int_0^1\frac{x^4\left(1-x^2\right)^5}{\left(1+x^2\right)^{10}}\, dx=A\]

Given the above, find $\frac{1}{A}.$

Main article: Taylor Series

Some functions like $\tfrac{1}{1-x}$, $\ln(1 - x)$, $\arctan x$, and $e^x$ have nice Taylor expansions that, together with term-by-term integration, can lead to a closed-form answer. The monotone convergence theorem states that in most cases where the integral does exist (as should generally be the case when evaluating an integral), the summation and integral may be interchanged. For more information, see Double Integrals .

Evaluate \[ \int_0^1 \ln x \ln(1 - x) \, dx.\] Note \[ \int_0^1 \ln x \ln(1-x) \, dx = - \int_0^1 \sum_{k = 1}^\infty \frac{x^k \ln x}{k} \, dx.\] Since the monotone convergence theorem applies here, this is equal to \[\begin{align} - \sum_{k = 1}^\infty \int_0^1 \frac{x^k \ln x}{k} \, dx &= \sum_{k = 1}^\infty \frac{1}{k (k+1)^2} \\ &= \sum_{k = 1}^\infty \frac{1}{k(k + 1)} - \sum_{k = 1}^\infty \frac{1}{(k + 1)^2} \\ &= 2 - \frac{\pi^2}{6}.\ _\square \end{align}\]

Main article: Differentiation under the integral sign

Differentiating under the integral sign is a useful method for evaluating certain integrals which might be harder using other methods. This method of integrating was so frequently used by Richard Feynman that it is often referred to as Feynman's integration trick .

\[\dfrac{d}{dx} \displaystyle \int_{g(x)}^{h(x)} f(x,t)dt = f\big(x,h(x)\big)\dfrac{d}{dx}h(x) - f\big(x,g(x)\big)\dfrac{d}{dx}g(x) + \displaystyle \int_{g(x)}^{h(x)}\dfrac{\partial}{\partial x}f(x,t)dt. \]

Compute \[ \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-7x}}{x} \, dx. \] Let $ I(a) = \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-ax}}{x}\, dx.$ Upon differentiating under the integral sign, the equation becomes \[\begin{align} \dfrac{\partial I}{\partial a} &= \displaystyle \int_{0}^{\infty} \dfrac{\partial }{\partial a}\dfrac{e^{-5x} - e^{-ax}}{x}\, dx \\ &= \displaystyle \int_{0}^{\infty} \dfrac{0 -(-xe^{-ax})}{x}\, dx \\ &= \displaystyle \int_{0}^{\infty} e^{-ax}\, dx. \end{align}\] Now, integrating with respect to $x$ yields the following: \[ \dfrac{\partial I}{\partial a} = \left[\dfrac{e^{-ax}}{a}\right]_{\infty}^{0} \implies \dfrac{\partial I}{\partial a} = \dfrac{1}{a} .\] Integrating both sides with respect to $a,$ \[ I(a) = \displaystyle \int \dfrac{1}{a}\, da + C \Rightarrow I(a) = \ln a + C, \qquad (1) \] where $C$ is the constant of integration. Notice that $ I(5) = \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-5x}}{x}\, dx = \int_{0}^{\infty} 0\, dx = 0. $ Substituting these values in $(1)$ gives \[ 0 = \ln 5 + C \implies C = -\ln 5 \implies I(a) = \ln\dfrac{a}{5}.\] To obtain the required integral, substitute $a = 7$: \[ I(7) = \ln\dfrac{7}{5} \implies \displaystyle \int_{0}^{\infty} \dfrac{e^{-5x} - e^{-7x}}{x}\, dx = \ln\dfrac{7}{5}. \ _\square\]

Compute \[ \displaystyle \int_{0}^{\infty} \dfrac{\sin x}{x}\, dx. \] Let $ I(a) = \displaystyle \int_{0}^{\infty} e^{-ax}\dfrac{\sin x}{x}\, dx. $ Then differentiating under the integral sign gives \[ \dfrac{\partial I(a)}{\partial a} = \displaystyle \int_{0}^{\infty} \dfrac{\partial}{\partial a} e^{-ax}\dfrac{\sin x}{x}\, dx= -\displaystyle \int_{0}^{\infty} e^{-ax}\sin x\, dx. \] Integrating with respect to $x,$ \[ \dfrac{\partial I(a)}{\partial a} = \left[\dfrac{e^{-ax}\left(a\sin x + \cos x\right)}{a^{2}+1}\right]_{0}^{\infty} = -\dfrac{1}{1+a^{2}}. \] Integrating with respect to $a,$ \[ I(a) = \displaystyle \int -\dfrac{1}{1+a^{2}}da = -\tan^{-1} a + C, \] where $C$ is the constant of integration. Now, \[ \displaystyle \lim_{a\rightarrow \infty} = \displaystyle \int_{0}^{\infty} \displaystyle \lim_{a\rightarrow \infty} e^{-ax}\dfrac{\sin x}{x}dx = 0 .\] Using the above information, \[ 0 = -\displaystyle \lim_{a \rightarrow \infty} \tan^{-1} a + C \Rightarrow C = \dfrac{\pi}{2} \Rightarrow I(a) = \dfrac{\pi}{2} - \tan^{-1} a. \] To get our integral, we let $a = 0$ to obtain \[ I(0) = \dfrac{\pi}{2} - \tan^{-1} 0 = \dfrac{\pi}{2}. \] Therefore, $ \displaystyle \int_{0}^{\infty} \dfrac{\sin x}{x}\, dx = \dfrac{\pi}{2} . \ _\square$

\[ \int_0^1 x^{50} (\ln x)^{150} \, dx \]

If the value of the integral above is equal to

\[ \dfrac{A!}{B^C}, \]

where $A,B,$ and $C$ are positive integers, find the value of $A+B+C$.

Bonus : Generalize $ \displaystyle \int_0^1 x^{m} (\ln x)^{n} \, dx $.

Find the value of the following integral:

\[ \displaystyle\int_{0}^{\infty} \dfrac{1}{x} \left(\tan^{-1} \pi x - \tan^{-1}x\right)\,dx.\]

Sometimes, the integrand looks like it has already been integrated. This may signal that the integral is better interpreted as a double integral . There are more possibilities for $u$-substitutions when two variables can be manipulated (polar, skewed, etc), and simply changing the order of integration may suffice to simplify the integral.

In many ways, this is a dual method to differentiation under the integral sign. The main difference is that the extra variable is interpreted inside of the variable rather than outside of it. In most cases where one works, both could work; in some cases, only one approach works nicely, so it is good to know both.

Suppose $a$ and $b$ are real numbers, $f$ a function, and $I = \int_a^b f(x) \, dx.$ If $f(x) = g(x, s) - g(x, r)$ for some constants $r$ and $ s$, then \[I = \int_a^b \int_r^s \frac{\partial}{\partial t} g(x,t) \, dt \, dx.\]

Supposing Fubini's theorem holds, the order of integration may be swapped or otherwise altered.

Evaluate \[\int_0^{\infty} \frac{e^{\pi x} - e^{x}}{x(e^{\pi x}+1)(e^{x}+1)} \, dx.\] We have \[ \begin{align*} \int_0^{\infty} \frac{e^{\pi x} - e^{x}}{x(e^{\pi x}+1)(e^{x}+1)} \, dx &= \int_0^\infty \frac{1}{x (e^x + 1)} - \frac{1}{x (e^{\pi x} + 1)} \, dx \\ &= \int_0^\infty \int_1^\pi \frac{e^{tx}}{(e^{tx} + 1)^2} \, dt \, dx \\ &= \int_1^\pi \int_0^\infty \frac{e^{tx}}{(e^{tx} + 1)^2} \, dx \, dt \\ &= \int_1^\pi \left[ - \frac{1}{t (e^{tx} + 1)} \right]_0^\infty \, dt \\ &= \int_1^\pi \frac{1}{(1 + 1)t} \, dt \\ &= \frac{\ln \pi}{2}.\ _\square \end{align*} \]

In complex analysis, a harmonic function is a real-valued function that is the real or imaginary part of a complex-differentiable function. In multivariable calculus, it is a function $f(x, \, y)$ such that $\left(\frac{\partial}{\partial x}\right)^2 f + \left( \frac{\partial}{\partial y}\right)^2f = 0.$ Generally, such facts from fields afar are not applicable to the evaluation of real integrals; however, the harmonic functions have a special property that greatly simplifies integration over circles .

Suppose $f$ is a bivariate harmonic function, $(a, \, b)$ is a point in the plane, and $r$ is a positive real number. Then, \[ \int_0^{2\pi} f(a + r\cos\theta, \, b + r\sin\theta) \, d\theta = 2\pi f(a, \, b).\]

Evaluate \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx.\] Consider the function $f(x, \, y) = e^x \cos y.$ Note that \[ \left(\frac{\partial}{\partial x}\right)^2 f + \left(\frac{\partial}{\partial y}\right)^2 f = e^x \cos y + e^x (- \cos y) = 0.\] Therefore, $f$ is a harmonic function, and it follows that \[\int_0^{2\pi} e^{\cos x} \cos(\sin x) \, dx = 2\pi e^0 \cos(0) = 2\pi.\ _\square\]

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Section 7.1 : Integration by Parts

Evaluate each of the following integrals.

INTEGRATION PRACTICE QUESTIONS WITH SOLUTIONS

Question 1 :

Integrate the following with respect to x

∫ (x + 5) 6 dx

∫ (x + 5) 6 dx = (x + 5) ( 6+1) /(6 +1) + c

= (x + 5) 7 /7 + c

Question 2 :

∫ 1/(2 - 3x) 4 dx

∫ 1/(2 - 3x) 4 dx = ∫ (2 - 3x) -4 dx

= (2 - 3x) (-4 + 1) / (-4 + 1) ⋅ (-3) + c

= (2 - 3x) -3 / (-3) (-3) + c

= (1/9) [1/ (2 - 3x) 3 ] + c

Question 3 :

∫ √(3x + 2) dx

∫ √(3x + 2) dx = ∫ (3x + 2) 1/2 dx

= (3x + 2) 3/2 / (3/2) (3) + c

= (3x + 2) 3/2 / (9/2) + c

= (2/9)(3x + 2) 3/2 + c

Question 4 :

∫ sin 3x dx

∫ sin 3x dx = (- cos 3x/3) + c

Question 5 :

∫ cos (5 - 11x) dx

∫ cos (5 - 11x) dx = sin (5 - 11x) / (-11) + c

= (-1/11) sin (5 - 11x) + c

Question 6 :

∫ cosec 2 (5x - 7) dx

∫ cosec 2 (5x - 7) dx = -cot (5x - 7) (1/5) + c

= (-1/5) cot (5x - 7) + c

Question 7 :

∫ e 3x- 6 dx

∫ e 3x- 6 dx = e 3x- 6 /3 + c

= (1/3) e 3x- 6 + c

Question 8 :

∫ e 8 - 7x dx

∫ e 8 - 7x dx = e 8 - 7x /(-7) + c

= (-1/7) e 8-7x + c

Question 9 :

∫ 1/(6 - 4x) dx

∫ 1/(6 - 4x) dx = (log (6 - 4x))/-4 + c

= (-1/4) (log (6 - 4x)) + c

Question 10 :

∫ sec 2 x/5 dx

∫ sec 2 x/5 dx = tan (x/5)/(1/5) + c

= 5 tan (x/5) + c

Question 11 :

∫ cosec (5x + 3) cot (5x + 3) dx

∫ cosec (5x + 3) cot (5x + 3) dx = [- cosec (5x + 3)]/5 + c

= (-1/5) cosec (5x + 3) + c

Question 12 :

∫ 30 sec (2 - 15x) tan (2 - 15x) dx

= 30 sec (2 - 15x)/(-15) + c

= -2 sec (2 - 15x) + c

Question 13 :

∫ 1/√(1 - (4x) 2 ) dx

∫ 1/ √(1 - (4x) 2 ) dx

= sin -1 (4x)/4 + c

= (1/4)sin -1 (4x) + c

Question 14 :

∫ 1/√(1 - 81x 2 ) dx

∫ 1/ √(1 - 81x 2 ) dx

= ∫ 1/ √(1 - (9x) 2 ) dx

= sin -1 (9x) / 9 + c

(1/9) sin -1 (9x) + c

Question 15 :

∫ 1/(1 + 36x 2 ) dx

= ∫ 1/(1 + (6x) 2 ) dx

= tan -1 (6x)/6 + c

= (1/6) tan -1 (6x) + c

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Maxima takes care of actually computing the integral of the mathematical function. Maxima's output is transformed to LaTeX again and is then presented to the user. The antiderivative is computed using the Risch algorithm , which is hard to understand for humans. That's why showing the steps of calculation is very challenging for integrals.

In order to show the steps, the calculator applies the same integration techniques that a human would apply. The program that does this has been developed over several years and is written in Maxima's own programming language. It consists of more than 17000 lines of code. When the integrand matches a known form, it applies fixed rules to solve the integral (e. g. partial fraction decomposition for rational functions, trigonometric substitution for integrands involving the square roots of a quadratic polynomial or integration by parts for products of certain functions). Otherwise, it tries different substitutions and transformations until either the integral is solved, time runs out or there is nothing left to try. The calculator lacks the mathematical intuition that is very useful for finding an antiderivative, but on the other hand it can try a large number of possibilities within a short amount of time. The step by step antiderivatives are often much shorter and more elegant than those found by Maxima.

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How to Integrate

Last Updated: March 3, 2023 References Approved

This article was co-authored by wikiHow staff writer, Hannah Madden . Hannah Madden is a writer, editor, and artist currently living in Portland, Oregon. In 2018, she graduated from Portland State University with a B.S. in Environmental Studies. Hannah enjoys writing articles about conservation, sustainability, and eco-friendly products. When she isn’t writing, you can find Hannah working on hand embroidery projects and listening to music. wikiHow marks an article as reader-approved once it receives enough positive feedback. In this case, several readers have written to tell us that this article was helpful to them, earning it our reader-approved status. This article has been viewed 261,570 times. Learn more...

Integration is the inverse operation of differentiation. It is commonly said that differentiation is a science, while integration is an art. The reason is because integration is simply a harder task to do - while a derivative is only concerned with the behavior of a function at a point, an integral, being a glorified sum, integration requires global knowledge of the function. So while there are some functions whose integrals can be evaluated using the standard techniques in this article, many more cannot.

We go over the basic techniques of single-variable integration in this article and apply them to functions with antiderivatives.

$\int _{{a}}^{{b}}f(x){\mathrm {d}}x$

Throughout this article, we will go over the process of finding antiderivatives of a function. An antiderivative is a function whose derivative is the original function we started with.

To verify that this power rule holds, differentiate the antiderivative to recover the original function.

$n=-1.$

$f(x)=x^{{4}}+2x^{{3}}-5x^{{2}}-1$

The common theme is that you must perform whatever manipulations in order to get the integral into a polynomial. From there, integration is easy. Judging whether the integral is easy enough to brute-force, or requires some algebraic manipulation first, is where the skill lies. [4] X Research source

Definite Integration

$\int _{{2}}^{{3}}x^{{2}}{\mathrm {d}}x$

This theorem is incredibly useful because it simplifies the integral and means that the definite integral is completely determined by just the values on its boundaries. There is no need to sum up rectangles anymore to compute integrals. All we need to do now is to find antiderivatives, and evaluate at the bounds!

${\begin{aligned}\int _{{2}}^{{3}}x^{{2}}{\mathrm {d}}x&={\frac {1}{3}}x^{{3}}{\Bigg |}_{{2}}^{{3}}\\&={\frac {1}{3}}(3)^{{3}}-{\frac {1}{3}}(2)^{{3}}\\&={\frac {19}{3}}\end{aligned}}$

Antiderivatives of Common Functions

$C,$

The reason for the absolute value in the logarithm function is subtle, and requires a more thorough understanding of real analysis in order to fully answer. For now, we will just live with the fact that the domains become the same when the absolute value bars are added.

$f(x)=2\cos x+\tan ^{{2}}x-6.$

Integrals of Symmetric Functions

$f(-x)=f(x).$

U-Substitution

1 See the main article on how to perform u-substitutions. U-substitution is a technique that changes variables with the hope of obtaining an easier integral. As we will see, it is the analogue of the chain rule for derivatives.

$e^{{ax}}$

Integration by Parts

$\int u{\mathrm {d}}v=uv-\int v{\mathrm {d}}u$

We converted the integral of a logarithm into the integral of 1, which is trivial to evaluate.

$\int \ln x{\mathrm {d}}x=x\ln x-x+C$

Community Q&A

This is what the diagram in step 2 actually shows - the blue rectangles are right-handed rectangles, the yellow are left-handed, the red sample the minimum on the interval, and the blue sample the maximum on the interval. The graph in the middle shows that all of these areas, when the number of rectangles is sent to infinity, converge to the integral.

About This Article

To integrate, you’ll typically be working with Riemann integrals, which is the summing up of rectangles. In order to sum up the area of rectangles, send the number of rectangles to infinity. As you increase the number of rectangles, the area of all the rectangles better approximates the area under the curve. The limit is what you’ll define as the integral of the function. This limit has to exist in order for the integral to have any meaning. To learn how to perform the power rule for integrals, keep reading! Did this summary help you? Yes No

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Frequently Asked Questions (FAQ)

What is the use of integration in real life.

Integrations is used in various fields such as engineering to determine the shape and size of strcutures. In Physics to find the centre of gravity. In the field of graphical representation to build three-dimensional models.

What is the best integral calculator?

Symbolab is the best integral calculator solving indefinite integrals, definite integrals, improper integrals, double integrals, triple integrals, multiple integrals, antiderivatives, and more.

What does to integrate mean?

Integration is a way to sum up parts to find the whole. It is used to find the area under a curve by slicing it to small rectangles and summing up thier areas.

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How to solve integration questions

Acquiring the tools for success, students must hone their skillset and know How to solve integration questions to stay competitive in today's educational environment.

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What are the best techniques in solving integration problems

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How to solve Integrals

Integrals theory.

Integrals are the sum of infinite summands, infinitely small.

Given a function f of a real variable x and an interval [a, b] of the real line, the definite integral

Bioprofe |Exams with exercises about physics, chemistry and mathematics | How to solve an integral | Integrals

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is defined informally to be the area of the region in the xy-plane bounded by the graph of f, the x-axis, and the vertical lines x = a and x = b, such that areas above the axis add to the total, and the area below the x axis subtract from the total.

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The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function f. In this case, it is called an indefinite integral and is written:

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KNOWN INTEGRALS

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INTEGRATION BY SUBSTITUTION

Whenever an integral can be written as:

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if we change t=u(x), the integral transforms in:

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may be easier to solve.

INTEGRATION BY PARTS

This method is useful in the cases where the integrating can put as the product of a function for the differential of other one

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INTEGRATION OF RATIONAL FUNCTIONS

A rational function is any function which can be written as the ratio of two polynomial functions.

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Proper: if the degree of the polynomial divisor is greater than the dividend.

Improper: if the dividend polynomial degree is greater than or equal to the divisor.

Any improper rational function can be decomposed into the sum of a polynomial plus a proper rational function.

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Therefore, the integral of an improper rational function can be written:

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To solve the integral of a rational function is decomposed into a sum of simple fractions:

1) The denominator is decomposed into a product of factors as follows:

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2) Is then written

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and then obtain the following expression:

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3) The coefficients A, B, …, N, are determined by successively x = a, x = b, etc.

For example:

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4) Coefficients obtained, we integrate expression.

Case where the denominator polynomial has multiple roots

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INTEGRATION BY TRIGONOMETRIC SUBSTITUTION

It is the substitution of trigonometric functions for other expressions. One may use the trigonometric identities to simplify certain integrals containing radical expressions.

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How to solve Derivatives

Calculation of a limit .

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5 Common Data Integration Challenges (And How to Solve Them)

Updated: April 21, 2021

Published: June 22, 2020

Most businesses handle a tremendous amount of data on a daily basis.

Internet users generate about 2.5 quintillion bytes of data each day , and more and more organizations are investing in big data and artificial intelligence (AI) . But how can your business make the most out of the data it has at its disposal?

Companies need to have a solid big data strategy to collect, process and analyze data. This is essential to make data-driven and informed decisions. A major component of an effective data strategy is data integration .

Learn More About HubSpot's Operations Hub Software

What's Data Integration?

Data integration means combining data from two or more disparate sources into one single source of truth. Throughout this process, data can be transformed or simply shared. Companies use data integration to obtain a unified view of their business. Depending on the business needs, you can integrate two sources completely or just certain types of data.

Integrating data enables organizations to access up-to-date, enriched, and valuable information about multiple areas in the business. Data about marketing performance, customer satisfaction, sales, and other processes become available across the company.

Simply put, data integration is essential to have a 360-degree view of your business and to empower you to make actual data-driven decisions. A 2019 survey found that 55% of data collected by companies is not used -- in other words, most companies have an untapped goldmine of data sitting in their systems.

With data integration, you can connect your software stack to ensure a continuous and effective data flow across your organization, making sure that all key players have access to the most relevant data when and where they most need it.

One of the best ways to achieve this is by employing integration software to combine the data within different applications in your software stack.

Data Integration Challenges

There are a multitude of solutions on the market to help you with this. However, even with so many resources available to create an amazing data integration strategy , there are still common mistakes to be avoided. Here's how to recognize and avoid them.

1. You have disparate data formats and sources

Your business is collecting data through a variety of applications, such as your accounting and billing software, lead generation tool, email marketing app, CRM, customer service application, and others.

Each one of these tools is accessed and maintained by different teams, and they each have their own processes for inputting and updating data. They might even be adding data into the system that already exists in other applications, or in different formats. For example, one team can be entering phone numbers into one application as (00) 555-5555, and another team is entering them in another application as +00 555 5555.

2. Your data isn't available where it needs to be

This results in your team wasting a lot of time and not having access to information that could make all the difference in the performance of their work -- which leads us to the second problem…

This is an issue that stems from the existence of data silos . Data silos are groups of data accessible by one department but isolated from the rest of the organization.

If there's no coherence as to how, who, and where to enter and update data, you inevitably end up with information silos across your organization.

Imagine that your marketing team is working on a new, highly personalized email campaign to your existing customers. As they're having discussions about how to collect customer data to create a more targeted campaign, your customer support team has been gathering exactly that kind of data -- and marketing knows nothing about it. The data sits inside your customer support software, while marketing is racking their brains trying to think of ways to get that information.

3. You have low-quality or outdated data

When you have no company-wide standards for data entry and maintenance - and when a lot of it still needs to be done manually -- you inevitably end up with inaccurate, outdated, and/or duplicate data.

Different departments might be inputting the same data into different systems, resulting in duplicates. Or, if your team needs to manually update data every so often, this can lead to mistakes in data entry or to huge amounts of data not being updated at all.

This can also happen if you go long periods of time without organizing your databases.

As a result, your data is inconsistent and untrustworthy -- and if you can't trust your data, you can't trust the analysis you get from it.

4. You're using the wrong integration software for your needs

Even if you're already using integration solutions to connect your software ecosystem, you can fall into the trap of using the wrong type of software for what you need -- or you might even have the right software, but you're using it the wrong way.

For example, you might be using a trigger-based integration to have the databases of two apps aligned. However, this solution doesn't sync historical data (data that was entered into your tools before the integration was set up) and it only pushes data from one platform into another. If what you want is for these databases to be synchronized, you'll need a two-way integration .

5. You have too much data

There is such a thing as too much data. If your company is collecting data indiscriminately, you end up with a lot of information you don't need, and it could be burying the valuable information beneath it. It's just like object hoarding: if your drawers are full of things you don't need, it makes it a whole lot harder to find the things you do need in the mess, and it takes you a lot more time to find it, too.

This problem is amplified if you're collecting data from multiple channels without a proper data management system in place. With the sheer amounts of data being created daily, it becomes a big challenge to manage, analyze, and extract value from your data when you can't find the signal in the noise.

How to Create a Top Data Integration Plan

If you're facing any or all of these challenges when coming up with a strategy to integrate your organization's data, fret not -- there are a few steps you can take to make sure your data integration plan goes smoothly.

1. Clean up your data

Cleaning up your data is an absolutely critical step to take before even thinking about integrating your software ecosystem. The first thing you need to do is to take a look at your existing databases and:

Cleaning up your databases might be time-consuming, but if you do it right and then set up an integration tool, you only need to do it once. Taking care of this will improve the quality of your data across the board and enable a seamless and effective integration process.

2. Introduce clear processes for data management

Next, introduce company-wide standards for data entry and maintenance. Data ownership is an important part of this: that means assigning one team or individual to be responsible for the quality and management of your data. They'll need to ensure that everything that enters your systems is up to code according to company policies and strategy.

If this isn't feasible for your organization, make sure to train all team members on how to correctly input and update data, as well as educating them on how your tools are connected.

By introducing company-wide protocols for data entry and management, you can significantly reduce the amount of low-quality, outdated or duplicate data in your system.

3. Back up your data

Before moving on to the actual data integration, an important -- and often overlooked — step is to back up your data . Your applications may already offer an option to back up your data, so check that this is possible with your software provider. You can back it up to the cloud or do a physical hard drive - or even both, if you want to be extra safe.

Once your data is cleaned up and backed up, you can move on to the actual integration.

4. Choose the right software to assist you with data integration

Having the right integration software to fit your needs is essential. It automates a huge part of your data management tasks and automatically syncs data between the applications in your software stack, drastically reducing the need for manual data entry, unifying data formats, and reducing the scope for error.

Integration software is the string that ties everything together in your stack. It ensures a continuous flow of data between different applications and makes sure that each team has access to the right information at the right time.

Before choosing an integration solution, you need to answer the following questions:

There are different kinds of integration platforms that work best for different use cases. You may want to use in-app integrations offered by the tools you're already using, or you may need to use a third-party integration platform or Integration Platform as a Service (iPaaS) provider.

Once you've assessed exactly what you need from an integration solution, you'll be in a much better position to choose the right software for you. If the tools you're using already offer native integrations that fulfill all of your requirements, that's a great starting point.

However, if those don't exactly fit your bill, deploying an iPaaS tool is a great idea. These tools include platforms like Zapier , Tray.io , and Automate.io , which specialize in automating workflows and one-way data pushes. These would allow you to create trigger-action workflows across your entire software stack.

If you need a continuous, real-time sync of your customer data, a tool like Operations Hub will probably work best for you. Operations Hub keeps your contacts flowing between your databases two ways and in real-time, which means that every time you change or update any contact data, that change is reflected in your other apps too. This also applies to data that was created before the sync was set up.

You might even decide that a mixture of integration software is the best for you, such as using Zapier to push some basic data one-way and Operations Hub to sync customer data across your tools.

5. Manage and maintain your data

These steps will help you automate huge parts of your data management strategy and make sure your organization has consistent, up-to-date and valuable data, which in turn will help you extract much better insights to make data-driven decisions.

But data management is an ongoing process: you still need to check in on your databases every so often to make sure everything is running smoothly, ensure that your team is following the correct processes, check if your existing tools are working as well as they should, and determine if any parts of your strategy need to be updated or adapted. This is especially important if your business is growing - your data integration strategy will likely need to evolve with you as you grow.

By being aware of these challenges and knowing how to tackle them in your data strategy, you'll be in the best position to collect and analyze the data your organization has at its disposal and make actionable, data-driven decisions.

Don't forget to share this post!

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Definite Integrals

You might like to read Introduction to Integration first!

Integration

The symbol for "Integral" is a stylish "S" (for "Sum", the idea of summing slices):

And then finish with dx to mean the slices go in the x direction (and approach zero in width).

Definite Integral

A Definite Integral has start and end values: in other words there is an interval [a, b].

a and b (called limits, bounds or boundaries) are put at the bottom and top of the "S", like this:

We find the Definite Integral by calculating the Indefinite Integral at a , and at b , then subtracting:

Example: What is 2 ∫ 1 2x dx

We are being asked for the Definite Integral , from 1 to 2, of 2x dx

First we need to find the Indefinite Integral .

Using the Rules of Integration we find that ∫ 2x dx = x 2 + C

Now calculate that at 1, and 2:

And "C" gets cancelled out ... so with Definite Integrals we can ignore C .

Check : with such a simple shape, let's also try calculating the area by geometry:

A = 2+4 2 × 1 = 3

Yes, it does have an area of 3.

Notation : It is usual to show the indefinite integral (without the +C) inside square brackets, with the limits a and b after, like this:

Example (continued)

How to show your answer:

Let's try another example:

The Definite Integral, from 0.5 to 1.0, of cos(x) dx:

(Note: x must be in radians )

The Indefinite Integral is: ∫ cos(x) dx = sin(x) + C

We can ignore C for definite integrals (as we saw above) and we get:

And another example to make an important point:

The Definite Integral, from 0 to 1, of sin(x) dx:

The Indefinite Integral is: ∫ sin(x) dx = −cos(x) + C

Since we are going from 0, can we just calculate the integral at x=1 ??

−cos(1) = −0.540...

What? It is negative ? But it looks positive in the graph.

Well ... we made a mistake !

Because we need to subtract the integral at x=0 . We shouldn't assume it is zero.

So let us do it properly, subtracting one from the other:

That's better!

But we can have negative regions , when the curve is below the axis:

The Definite Integral, from 1 to 3, of cos(x) dx:

Notice that some of it is positive, and some negative. The definite integral will work out the net value.

Let us do the calculations:

So there is more negative than positive with a net result of −0.700....

Try integrating cos(x) with different start and end values to see for yourself how positives and negatives work.

Positive Area

But sometimes we want all area treated as positive (without the part below the axis being subtracted).

In that case we must calculate the areas separately , like in this example:

Example: What is the total area between y = cos(x) and the x-axis, from x = 1 to x = 3?

This is like the example we just did, but now we expect that it is all positive (imagine we had to paint it).

So now we have to do the parts separately:

The curve crosses the x-axis at x = π /2 so we have:

From 1 to π /2:

From π /2 to 3:

That last one comes out negative, but we want it to be positive, so:

Total area = 0.158... + 0.859... = 1.017 ...

This is very different from the answer in the previous example.

Oh yes, the function we are integrating must be Continuous between a and b : no holes, jumps or vertical asymptotes (where the function heads up/down towards infinity).

A vertical asymptote between a and b affects the definite integral.

Area above − area below

The integral adds the area above the axis but subtracts the area below, for a "net value":

Adding Functions

The integral of f+g equals the integral of f plus the integral of g :

Reversing the interval

Reversing the direction of the interval gives the negative of the original direction.

Interval of zero length

When the interval starts and ends at the same place, the result is zero:

Adding intervals

We can also add two adjacent intervals together:

The Definite Integral between a and b is the Indefinite Integral at b minus the Indefinite Integral at a .

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5.7: Integrals Resulting in Inverse Trigonometric Functions

Gilbert Strang & Edwin “Jed” Herman

Learning Objectives

Integrate functions resulting in inverse trigonometric functions

In this section we focus on integrals that result in inverse trigonometric functions. We have worked with these functions before. Recall, that trigonometric functions are not one-to-one unless the domains are restricted. When working with inverses of trigonometric functions, we always need to be careful to take these restrictions into account. Also, we previously developed formulas for derivatives of inverse trigonometric functions. The formulas developed there give rise directly to integration formulas involving inverse trigonometric functions.

Integrals that Result in Inverse Trigonometric Functions

Let us begin this last section of the chapter with the three formulas. Along with these formulas, we use substitution to evaluate the integrals. We prove the formula for the inverse sine integral.

Rule: Integration Formulas Resulting in Inverse Trigonometric Functions

The following integration formulas yield inverse trigonometric functions:

\[ \begin{align} ∫\dfrac{du}{\sqrt{a^2−u^2}} =\sin^{−1}\left(\dfrac{u}{a}\right)+C \\ ∫\dfrac{du}{a^2+u^2} =\dfrac{1}{a}\tan^{−1}\left(\dfrac{u}{a}\right)+C \\ ∫\dfrac{du}{u\sqrt{u^2−a^2}} =\dfrac{1}{a}\sec^{−1}\left(\dfrac{|u|}{a}\right)+C \end{align} \nonumber \]

Proof of the first formula

Let $ y=\sin^{−1}\frac{x}{a}$. Then $ a \sin y=x$. Now using implicit differentiation, we obtain

\[ \dfrac{d}{dx}(a \sin y)=\dfrac{d}{dx}(x) \nonumber \]

\[ a\cos y\dfrac{dy}{dx}=1 \nonumber \]

\[ \dfrac{dy}{dx}=\dfrac{1}{a\cos y}. \nonumber \]

For $ −\dfrac{π}{2}≤y≤\dfrac{π}{2},\cos y≥0.$ Thus, applying the Pythagorean identity $ \sin^2y+\cos^2y=1$, we have $ \cos y=\sqrt{1-\sin^2y}.$ This gives

\[ \begin{align} \dfrac{1}{a \cos y} =\dfrac{1}{a\sqrt{1−\sin^2y}} \\ =\dfrac{1}{\sqrt{a^2−a^2 \sin^2y}} \\ =\dfrac{1}{\sqrt{a^2−x^2}}. \end{align} \nonumber \]

Then for $ −a≤x≤a,$ we have

\[ ∫\dfrac{1}{\sqrt{a^2−u^2}}\,du=\sin^{−1}\left(\frac{u}{a}\right)+C. \nonumber \]

Example $ \PageIndex{1}$: Evaluating a Definite Integral Using Inverse Trigonometric Functions

Evaluate the definite integral

\[ ∫^{1/2}_0\dfrac{dx}{\sqrt{1−x^2}}. \nonumber \]

We can go directly to the formula for the antiderivative in the rule on integration formulas resulting in inverse trigonometric functions, and then evaluate the definite integral. We have

\[\int_0^{1/2}\dfrac{dx}{\sqrt{1-x^2}} = \sin^{-1} x \,\bigg|_0^{1/2} = \sin^{-1} \tfrac{1}{2} - \sin^{-1} 0 = \dfrac{\pi}{6}-0 = \dfrac{\pi}{6}. \nonumber \]

Note that since the integrand is simply the derivative of $\sin^{-1} x$, we are really just using this fact to find the antiderivative here.

Exercise $\PageIndex{1}$

Find the indefinite integral using an inverse trigonometric function and substitution for $\displaystyle ∫\dfrac{dx}{\sqrt{9−x^2}}$.

Use the formula in the rule on integration formulas resulting in inverse trigonometric functions.

$ \displaystyle ∫\dfrac{dx}{\sqrt{9−x^2}} \quad=\quad \sin^{−1}\left(\dfrac{x}{3}\right)+C $

In many integrals that result in inverse trigonometric functions in the antiderivative, we may need to use substitution to see how to use the integration formulas provided above.

Example $ \PageIndex{2}$: Finding an Antiderivative Involving an Inverse Trigonometric Function using substitution

Evaluate the integral

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}.\nonumber \]

Substitute $ u=3x$. Then $ du=3\,dx$ and we have

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}.\nonumber \]

Applying the formula with $ a=2,$ we obtain

\[ ∫\dfrac{dx}{\sqrt{4−9x^2}}=\dfrac{1}{3}∫\dfrac{du}{\sqrt{4−u^2}}=\dfrac{1}{3}\sin^{−1}\left(\dfrac{u}{2}\right)+C=\dfrac{1}{3}\sin^{−1}\left(\dfrac{3x}{2}\right)+C.\nonumber \]

Exercise $\PageIndex{2}$

Find the antiderivative of $\displaystyle ∫\dfrac{dx}{\sqrt{1−16x^2}}.$

Substitute $ u=4x$.

$ \displaystyle ∫\dfrac{dx}{\sqrt{1−16x^2}} = \dfrac{1}{4}\sin^{−1}(4x)+C$

Example $ \PageIndex{3}$: Evaluating a Definite Integral

\[ ∫^{\sqrt{3}/2}_0\dfrac{du}{\sqrt{1−u^2}}\nonumber. \nonumber \]

The format of the problem matches the inverse sine formula. Thus,

\[ ∫^{\sqrt{3}/2}_0\dfrac{du}{\sqrt{1−u^2}}=\sin^{−1}u\,\bigg|^{\sqrt{3}/2}_0=[\sin^{−1}\left(\dfrac{\sqrt{3}}{2}\right)]−[\sin^{−1}(0)]=\dfrac{π}{3}.\nonumber \]

Integrals Resulting in Other Inverse Trigonometric Functions

There are six inverse trigonometric functions. However, only three integration formulas are noted in the rule on integration formulas resulting in inverse trigonometric functions because the remaining three are negative versions of the ones we use. The only difference is whether the integrand is positive or negative. Rather than memorizing three more formulas, if the integrand is negative, simply factor out −1 and evaluate the integral using one of the formulas already provided. To close this section, we examine one more formula: the integral resulting in the inverse tangent function.

Example $ \PageIndex{4}$: Finding an Antiderivative Involving the Inverse Tangent Function

Find the antiderivative of $\displaystyle ∫\dfrac{1}{9+x^2}\,dx.$

Apply the formula with $ a=3$. Then,

\[ ∫\dfrac{dx}{9+x^2}=\dfrac{1}{3}\tan^{−1}\left(\dfrac{x}{3}\right)+C. \nonumber \]

Exercise $\PageIndex{3}$

Find the antiderivative of $\displaystyle ∫\dfrac{dx}{16+x^2}$.

Follow the steps in Example $ \PageIndex{4}$.

$\displaystyle ∫\dfrac{dx}{16+x^2} = \frac{1}{4}\tan^{−1}\left(\dfrac{x}{4}\right)+C $

Example $ \PageIndex{5}$: Applying the Integration Formulas WITH SUBSTITUTION

Find an antiderivative of $\displaystyle ∫\dfrac{1}{1+4x^2}\,dx.$

Comparing this problem with the formulas stated in the rule on integration formulas resulting in inverse trigonometric functions, the integrand looks similar to the formula for $ \tan^{−1} u+C$. So we use substitution, letting $ u=2x$, then $ du=2\,dx$ and $ \dfrac{1}{2}\,du=dx.$Then, we have

\[ \dfrac{1}{2}∫\dfrac{1}{1+u^2}\,du=\dfrac{1}{2}\tan^{−1}u+C=\dfrac{1}{2}\tan^{−1}(2x)+C. \nonumber \]

Exercise $\PageIndex{4}$

Use substitution to find the antiderivative of $\displaystyle ∫\dfrac{dx}{25+4x^2}.$

Use the solving strategy from Example $ \PageIndex{5}$ and the rule on integration formulas resulting in inverse trigonometric functions.

$\displaystyle ∫\dfrac{dx}{25+4x^2} = \dfrac{1}{10}\tan^{−1}\left(\dfrac{2x}{5}\right)+C $

Example $ \PageIndex{6}$: Evaluating a Definite Integral

Evaluate the definite integral $\displaystyle ∫^{\sqrt{3}}_{\sqrt{3}/3}\dfrac{dx}{1+x^2}$.

Use the formula for the inverse tangent. We have

\[\int_{\sqrt{3}/3}^{\sqrt{3}} \frac{dx}{1+x^2} = \tan^{-1}x\,\bigg|_{\sqrt{3}/3}^{\sqrt{3}} = [\tan^{-1}\left(\sqrt{3}\right)] - [\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)] =\frac{\pi}{3} - \frac{\pi}{6} =\frac{\pi}{6}.\nonumber \]

Exercise $\PageIndex{5}$

Evaluate the definite integral $\displaystyle ∫^2_0\dfrac{dx}{4+x^2}$.

Follow the procedures from Example $\PageIndex{6}$ to solve the problem.

$\displaystyle ∫^2_0\dfrac{dx}{4+x^2} = \dfrac{π}{8} $

Key Concepts

Key Equations

Integrals That Produce Inverse Trigonometric Functions

$\displaystyle ∫\dfrac{du}{\sqrt{a^2−u^2}}=\sin^{−1}\left(\dfrac{u}{a}\right)+C$

$\displaystyle ∫\dfrac{du}{a^2+u^2}=\dfrac{1}{a}\tan^{−1}\left(\dfrac{u}{a}\right)+C$

$\displaystyle ∫\dfrac{du}{u\sqrt{u^2−a^2}}=\dfrac{1}{a}\sec^{−1}\left(\dfrac{|u|}{a}\right)+C$

Contributors and Attributions

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org .

Includes some added textual clarifications and edits by Paul Seeburger (Monroe Community College)

Math Article

Integration

Integration is the calculation of an integral. Integrals in maths are used to find many useful quantities such as areas, volumes, displacement, etc. When we speak about integrals, it is related to usually definite integrals. The indefinite integrals are used for antiderivatives. Integration is one of the two major calculus topics in Mathematics, apart from differentiation(which measure the rate of change of any function with respect to its variables). It’s a vast topic which is discussed at higher level classes like in Class 11 and 12. Integration by parts and by the substitution is explained broadly. Here, you will learn the definition of integrals in Maths, formulas of integration along with examples.

Table of Contents:

Integration Definition

Integral calculus, integration – inverse process of differentiation, definite integral, indefinite integral, integration formulas.

The integration denotes the summation of discrete data. The integral is calculated to find the functions which will describe the area, displacement, volume, that occurs due to a collection of small data, which cannot be measured singularly. In a broad sense, in calculus, the idea of limit is used where algebra and geometry are implemented. Limits help us in the study of the result of points on a graph such as how they get closer to each other until their distance is almost zero . We know that there are two major types of calculus –

Differential Calculus

The concept of integration has developed to solve the following types of problems:

These two problems lead to the development of the concept called the “Integral Calculus”, which consist of definite and indefinite integral. In calculus, the concept of differentiating a function and integrating a function is linked using the theorem called the Fundamental Theorem of Calculus.

Maths Integration

In Maths, integration is a method of adding or summing up the parts to find the whole. It is a reverse process of differentiation, where we reduce the functions into parts. This method is used to find the summation under a vast scale. Calculation of small addition problems is an easy task which we can do manually or by using calculators as well. But for big addition problems, where the limits could reach to even infinity, integration methods are used. Integration and differentiation both are important parts of calculus. The concept level of these topics is very high. Hence, it is introduced to us at higher secondary classes and then in engineering or higher education. To get an in-depth knowledge of integrals, read the complete article here.

According to Mathematician Bernhard Riemann,

“Integral is based on a limiting procedure which approximates the area of a curvilinear region by breaking the region into thin vertical slabs.” Learn more about Integral calculus here.

Let us now try to understand what does that mean:

Take an example of a slope of a line in a graph to see what differential calculus is:

In general, we can find the slope by using the slope formula. But what if we are given to find an area of a curve? For a curve, the slope of the points varies, and it is then we need differential calculus to find the slope of a curve.

You must be familiar with finding out the derivative of a function using the rules of the derivative. Wasn’t it interesting? Now you are going to learn the other way round to find the original function using the rules in Integrating.

We know that differentiation is the process of finding the derivative of the functions and integration is the process of finding the antiderivative of a function. So, these processes are inverse of each other. So we can say that integration is the inverse process of differentiation or vice versa. The integration is also called the anti-differentiation. In this process, we are provided with the derivative of a function and asked to find out the function (i.e., primitive).

We know that the differentiation of sin x is cos x.

It is mathematically written as:

(d/dx) sinx = cos x …(1)

Here, cos x is the derivative of sin x. So, sin x is the antiderivative of the function cos x. Also, any real number “C” is considered as a constant function and the derivative of the constant function is zero.

So, equation (1) can be written as

(d/dx) (sinx + C)= cos x +0

(d/dx) (sinx + C)= cos x

Where “C” is the arbitrary constant or constant of integration.

Generally, we can write the function as follow:

(d/dx) [F(x)+C] = f(x), where x belongs to the interval I.

To represent the antiderivative of “f”, the integral symbol “∫” symbol is introduced. The antiderivative of the function is represented as ∫ f(x) dx. This can also be read as the indefinite integral of the function “f” with respect to x.

Therefore, the symbolic representation of the antiderivative of a function (Integration) is:

y = ∫ f(x) dx

∫ f(x) dx = F(x) + C.

Integrals in Maths

You have learned until now the concept of integration. You will come across, two types of integrals in maths:

An integral that contains the upper and lower limits then it is a definite integral. On a real line, x is restricted to lie. Riemann Integral is the other name of the Definite Integral.

A definite Integral is represented as:

$\begin{array}{l}\int_{a}^{b} f(x)dx\end{array} $

Indefinite integrals are defined without upper and lower limits. It is represented as:

∫f(x)dx = F(x) + C

Where C is any constant and the function f(x) is called the integrand.

Check below the formulas of integral or integration, which are commonly used in higher-level maths calculations. Using these formulas, you can easily solve any problems related to integration.

Also, get some more complete definite integral formulas here.

Integration Examples

Solve some problems based on integration concept and formulas here.

Example 1: Find the integral of the function: $\begin{array}{l}\int_{0}^{3} x^{2}dx\end{array} $

$\begin{array}{l}Given:\ \int_{0}^{3} x^{2}dx\end{array} $

$\begin{array}{l} = \left ( \frac{x^{3}}{3} \right )_{0}^{3}\end{array} $

$\begin{array}{l}= \left ( \frac{3^{3}}{3} \right ) – \left ( \frac{0^{3}}{3} \right )\end{array} $

Example 2: Find the integral of the function: ∫x 2 dx

Given ∫x 2 dx

= (x 3 /3) + C.

Integrate ∫ (x 2 -1)(4+3x)dx.

Given: ∫ (x 2 -1)(4+3x)dx.

Multiply the terms, we get

∫ (x 2 -1)(4+3x)dx = ∫ 4x 2 +3x 3 -3x-4 dx

Now, integrate it, we get

∫ (x 2 -1)(4+3x)dx = 4(x 3 /3) + 3(x 4 /4)- 3(x 2 /2) – 4x + C

The antiderivative of the given function ∫ (x 2 -1)(4+3x)dx is 4(x 3 /3) + 3(x 4 /4)- 3(x 2 /2) – 4x + C.

Video Lesson on Class 12 Important Calculus Questions

Frequently Asked Questions on Integration

What is integration.

The integration is the process of finding the antiderivative of a function. It is a similar way to add the slices to make it whole. The integration is the inverse process of differentiation.

What is the use of integration?

The integration is used to find the volume, area and the central values of many things.

What are the real-life applications of integration?

Integrations are much needed to calculate the centre of gravity, centre of mass, and helps to predict the position of the planets, and so on.

What is the fundamental theorem of calculus?

The fundamental theorem of calculus links the concept of differentiation and integration of a function.

Mention two different types of integrals in Maths.

Integration is one of the two main concepts of Maths, and the integral assigns a number to the function. The two different types of integrals are definite integral and indefinite integral.

Download BYJU’S – The Learning App to get personalised videos for all the important Maths topics. Also, learn about differentiation-integration concepts briefly here.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

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Integration

Integration is the inverse of differentiation of algebraic and trigonometric expressions involving brackets and powers. This can solve differential equations and evaluate definite integrals.

Read more about sharing

Integrating simple algebraic expressions

Integration is the inverse process to differentiation. Some people call it anti-differentiation .

Instead of multiplying the power at the front and subtracting one from the power, we add one to the power and then divide by the new power.

\[\int {{x^2}}\,\, dx\]

This just means, integrate ${x^2}$ with respect to $x$ . Remember, add one to the power and divide by the new power.

\[= \frac{{{x^3}}}{3} + c\]

The $+ c$ appears because when you differentiate a constant term, the answer is zero, so as we are performing 'anti-differentiation', we presume there may have been a constant term, which reduced to zero when differentiated. This $c$ is called the constant of integration.

In general:

$\frac{{dy}}{{dx}} = a{x^n} \to y = \frac{{a{x^{n + 1}}}}{{n + 1}} + c$ provided $n \ne - 1$

In other words you add one to the power, divide by the new power and add the constant of integration.

Find $\int {({x^4}} + {x^3})\,\,dx$

\[\int {({x^4}} + {x^3})\,\,dx\]

\[= \frac{{{x^5}}}{5} + \frac{{{x^4}}}{4} + c\]

Find $\int {(4{x^3}} + 7{x^{ - 2}})\,\,dx$

\[\int {(4{x^3}} + 7{x^{ - 2}})\,\,dx\]

\[= \frac{{4{x^4}}}{4} + \frac{{7{x^{ - 1}}}}{{ - 1}} + c\]

\[= {x^4} - \frac{7}{x} + c\]

Find $\int {{{(x + 2)}^2}}\,\,dx$

\[\int {{{(x + 2)}^2}}\,\, dx\]

\[= \int {({x^2}}+ 4x + 4)\,\,dx\]

\[= \frac{{{x^3}}}{3} + \frac{{4{x^2}}}{2} + 4x + c\]

\[= \frac{{{x^3}}}{3} + 2{x^2} + 4x + c\]

Find $\int {\frac{{x + \sqrt x + \sqrt[3]{x}}}{x}}\,\,dx$

\[\int {\frac{{x + \sqrt x + \sqrt[3]{x}}}{x}}\,\,dx\]

\[= \int {\frac{{x + {x^{\frac{1}{2}}} + {x^{\frac{1}{3}}}}}{x}}\,\,dx\]

\[= \int {\frac{x}{x}} + \frac{{{x^{\frac{1}{2}}}}}{x} + \frac{{{x^{\frac{1}{3}}}}}{x}\,\,dx\]

\[= \int {(1 + {x^{ - \frac{1}{2}}}} + {x^{ - \frac{2}{3}}})\,\,dx\]

\[= x + \frac{{{x^{\frac{1}{2}}}}}{{\frac{1}{2}}} + \frac{{{x^{\frac{1}{3}}}}}{{\frac{1}{3}}} + c\]

\[= x + 2\sqrt x + 3\sqrt[3]{x} + c\]

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