Integration by Substitution

Integration by Substitution

Theorem 3..4 Let be a continuous function. If $x = \phi(t)$ and $\phi(t)$ is differentiable, then

$\displaystyle \int f(x)dx = \int f(\phi(t))\phi^{\prime}(t)dt$

NOTE Let $x = \phi(t)$ . Then $dx = \frac{dx}{dt}dt = \phi'(t)dt$ . If the integrand of a given integral is transformed to the known form of the Rules of integration, then we can solve the problem by integration by substitution..

Example 3..3 Let . Integrate the following functions
1. $\displaystyle{\int \frac{2x + 1}{x^2 + x} dx}$ 2. $\displaystyle{\frac{x}{\sqrt{x^2-4}}}$ 3. $\displaystyle{\frac{1}{e^{x} + e^{-x}}\:dx}$
SOLUTION 1. Let . Then $dt = \frac{dt}{dx}dx = \frac{d(x^2+x)}{dx}dx = (2x + 1)dx$ . Thus,

$\displaystyle \int \frac{2x + 1}{x^2 + x} dx = \int \frac{dt}{t}.$
By the integration,

$\displaystyle \int \frac{2x + 1}{x^2 + x} dx = \int \frac{dt}{t} = \log{\vert t\vert} + c = \log{\vert x^2 + x\vert} + c\ensuremath{\ \blacksquare}$

2. Let . Then .

$\displaystyle \int \frac{x}{\sqrt{x^2 - 4}}dx$ $\displaystyle =$ $\displaystyle \int \frac{dt/2}{\sqrt{t}} = \frac{1}{2}\int t^{-1/2}dt$

$\displaystyle =$ $\displaystyle \frac{1}{2}2t^{1/2} + c = \sqrt{x^2 - 4} + c\ensuremath{\ \blacksquare}$

3.

$\displaystyle \int \frac{1}{e^{x} + e^{-x}}\:dx$ $\displaystyle =$ $\displaystyle \int \frac{1}{e^{x} + \frac{1}{e^{x}}}\:dx= \int \frac{e^{x}}{e^{2x} + 1}\:dx$

Now let $t = e^{x}$ . Then $dt = e^{x}dx$ . Thus

$\displaystyle \int \frac{e^{x}}{e^{2x} + 1}\:dx$ $\displaystyle =$ $\displaystyle \int \frac{dt}{t^2 + 1}\:dx$

$\displaystyle =$ $\displaystyle \tan^{-1}{t} + c = \tan^{-1}(e^{x}) + c\ensuremath{\ \blacksquare}$

Exercise 3..3 Let . Then integrate the following functions.
1. $\displaystyle{\int x^{2}\sqrt{x+1}dx}$ 2. $\displaystyle{\int \frac{1}{\sqrt{a^2 - x^2}}}$
SOLUTION 1. Let $t = \sqrt{x+1}$ . Then and . Thus

$\displaystyle \int x^{2}\sqrt{x+1}dx$ $\displaystyle =$ $\displaystyle \int (t^2 - 1)^2(t)(2t)dt$

$\displaystyle =$ $\displaystyle \int 2(t^{4} -2t^2 + 1)t^{2}dt$

$\displaystyle =$ $\displaystyle 2 \int (t^{6} - 2t^{4} + t^{2})dt$

$\displaystyle =$ $\displaystyle 2\big(\frac{1}{7}t^{7} - \frac{2}{5}t^{5} + \frac{1}{3}t^{3}\big) + c\$

$\displaystyle =$ $\displaystyle \frac{2}{7}(x+1)^{\frac{7}{2}} - \frac{4}{5}(x+1)^{\frac{5}{2}} + \frac{2}{3}(x+1)^{\frac{3}{2}} + c\ensuremath{\ \blacksquare}$

2. Let $x = a\sin{t}$ where $-\frac{\pi}{2} \leq t \leq \frac{\pi}{2})$ . Then $dx = a\cos{t}dt$ . Since $\sqrt{a^2 -x^2} = \sqrt{a^2 - a^2 \sin^{2}{t}} = \sqrt{a^2(1 - \sin^{2}{t})} = \sqrt{a^2 \cos^2{t}} = a\cos{t}$ , we have

$\displaystyle \int \frac{1}{\sqrt{a^2 -x^2}}\;dx = \int \frac{a\cos{t}}{a\cos{t}}\;dt = t + c.$
Note that $t = \sin^{-1}{\frac{x}{a}}$ . Thus

$\displaystyle \int \frac{1}{\sqrt{a^2 -x^2}}\;dx = \sin^{-1}{\frac{x}{a}} + c\ensuremath{\ \blacksquare}$

Exercise A

1.

Work out the following integrals?D
(a) $\displaystyle{\int{\sin{2x}}\ dx}$ (b) $\displaystyle{\int{\frac{x}{x^{2}+1}}\ dx}$ (c) $\displaystyle{\int{e^{2x}}\ dx}$ (d) $\displaystyle{\int{\frac{1}{x\log{x}}}\ dx}$
(e) $\displaystyle{\int{xe^{x^{2}}}\ dx}$ (f) $\displaystyle{\int{\sin^{2}{x}\cos{x}}\ dx}$ (g) $\displaystyle{\int{x\sqrt{1+x}} \ dx}$
(h) $\displaystyle{\int{\frac{\cos{x} - \sin{x}}{\sin{x} + \cos{x}}}\ dx }$ (i) $\displaystyle{\int{\frac{e^x}{1 + e^{x}}} \ dx}$

Exercise B

1.

Work out the following integrals?D
(a) $\displaystyle{\int{e^{2-x}}\ dx}$ (b) $\displaystyle{\int{\sec^2{(1-x)}}\ dx}$ (c) $\displaystyle{\int{\frac{x}{\sqrt{1-x^2}}}\ dx}$ (d) $\displaystyle{\int{\frac{\sin{x}}{\cos^2{x}}}\ dx}$ (e) $\displaystyle{\int{\frac{e^{1/x}}{x^2}} \ dx}$ (f) $\displaystyle{\int{\frac{\sec^2{\theta}}{\sqrt{3\tan{\theta} + 1}}}\ d\theta}$ (g) $\displaystyle{\int{\frac{1+\cos{2x}}{\sin^2{x}}} \ dx}$ (h) $\displaystyle{\int{\frac{\log{x}}{x}}\ dx }$ (i) $\displaystyle{\int{\frac{e^x}{1 + e^{2x}}} \ dx}$ (j) $\displaystyle{\int{x\sin{(x^2)}} \ dx}$

Next: Integration by Parts Up: Integration Previous: Antiderivatives Contents Index

$\displaystyle \int \frac{x}{\sqrt{x^2 - 4}}dx$	$\displaystyle =$	$\displaystyle \int \frac{dt/2}{\sqrt{t}} = \frac{1}{2}\int t^{-1/2}dt$
	$\displaystyle =$	$\displaystyle \frac{1}{2}2t^{1/2} + c = \sqrt{x^2 - 4} + c\ensuremath{\ \blacksquare}$

$\displaystyle \int \frac{e^{x}}{e^{2x} + 1}\:dx$	$\displaystyle =$	$\displaystyle \int \frac{dt}{t^2 + 1}\:dx$
	$\displaystyle =$	$\displaystyle \tan^{-1}{t} + c = \tan^{-1}(e^{x}) + c\ensuremath{\ \blacksquare}$

$\displaystyle \int x^{2}\sqrt{x+1}dx$	$\displaystyle =$	$\displaystyle \int (t^2 - 1)^2(t)(2t)dt$
	$\displaystyle =$	$\displaystyle \int 2(t^{4} -2t^2 + 1)t^{2}dt$
	$\displaystyle =$	$\displaystyle 2 \int (t^{6} - 2t^{4} + t^{2})dt$
	$\displaystyle =$	$\displaystyle 2\big(\frac{1}{7}t^{7} - \frac{2}{5}t^{5} + \frac{1}{3}t^{3}\big) + c\$
	$\displaystyle =$	$\displaystyle \frac{2}{7}(x+1)^{\frac{7}{2}} - \frac{4}{5}(x+1)^{\frac{5}{2}} + \frac{2}{3}(x+1)^{\frac{3}{2}} + c\ensuremath{\ \blacksquare}$