Integration by Substitution

Integration by Substitution

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 $a > 0$

. Integrate the following functions
1. $\displaystyle{\int \frac{2x + 1}{x^2 + x} dx}$ 2. $\displaystyle{\frac{x}{\sqrt{x^2-4}}}$ 3. $\displaystyle{\frac{1}{e^{x} + e^{-x}}\:dx}$

We let $u = $ the denominator. SOLUTION 1. Let $t = x^2 + x$ . 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,

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

For $\frac{1st degree polynomial}{\sqrt{2nd degree polynomial}}$ , let $t =$ be 2nd degree polynomial. Then $dt = 2xdx$ and we can express the integrand and $dx$ as a function of $t$ and $dt$ .

Let $e^{-x} = \frac{1}{e^{x}}$ . Then $e^{x}e^{x} = e^{x+x} = e^{2x}$ .

2. Let $t = x^2 - 4$ . Then $dt = 2xdx$ .

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

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

$\displaystyle =$

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

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 $a > 0$

. Then integrate the following functions.
1. $\displaystyle{\int x^{2}\sqrt{x+1}dx}$ 2. $\displaystyle{\int \frac{1}{\sqrt{a^2 - x^2}}}$

For $\sqrt{\rm 1st degree polynomial}$ , let $t = \sqrt{\rm 1st degree polynomial}$ .

SOLUTION 1. Let $t = \sqrt{x+1}$ . Then $t^2 = x+1$ and $2tdt = dx$ . Thus

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

Note that the integrand contains $a^2 - x^2$ . Then consider the following right triangle.

**Figure 3.1:** Exercise3-3-2
$\includegraphics[width=3cm]{SOFTFIG-3/trigsub1.eps}$

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

(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
(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) $\displaystyle{\int{x\sin{(x^2)}} dx}$