Integration by Substitution

Integration by Substitution

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

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

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. $${\int \frac{2x + 1}{x^2 + x} dx}$$2. $${\frac{x}{\sqrt{x^2-4}}}$$ 3. $${\frac{1}{e^{x} + e^{-x}}\:dx}$$

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,

$$\int \frac{2x + 1}{x^2 + x} dx = \int \frac{dt}{t}.$$

By the integration,

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

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 $dt = 2xdx$.

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

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

3.

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

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

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

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

Exercise 3..3   Let $a > 0$. Then integrate the following functions.
1. $${\int x^{2}\sqrt{x+1}dx}$$ 2. $${\int \frac{1}{\sqrt{a^2 - x^2}}}$$

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

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

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

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

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

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

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

Figure 3.1: Exercise3-3-2

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

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

Note that $t = \sin^{-1}{\frac{x}{a}}$. Thus

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

Exercise A

1.

Work out the following integrals

(a) $${\int{\sin{2x}} dx}$$ (b) $${\int{\frac{x}{x^{2}+1}} dx}$$ (c) $${\int{e^{2x}} dx}$$ (d) $${\int{\frac{1}{x\log{x}}} dx}$$

(e) $${\int{xe^{x^{2}}} dx}$$ (f) $${\int{\sin^{2}{x}\cos{x}} dx}$$ (g) $${\int{x\sqrt{1+x}} dx}$$

(h) $${\int{\frac{\cos{x} - \sin{x}}{\sin{x} + \cos{x}}} dx }$$ (i) $${\int{\frac{e^x}{1 + e^{x}}} dx}$$

Exercise B

1.

Work out the following integrals

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