Evaluating Definite Integral

u-substitution

Theorem 3..11 If $x = \phi(u)$ is differentiable on the open interval and is continuous on the closed interval $[\phi(a), \phi(b)]$ , then

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

NOTE Let $x = \phi(u)$ . Then $dx = \phi'(u)du$ . Now the limit of intervals must be changed from $x = \phi(a)$ to and $x = \phi(b)$ to .
$\begin{displaymath}\begin{array}{l\vert lll} x & \phi(a) & \to & \phi(b)\\ \hline t & a & \to & b \end{array}\end{displaymath}$ .

Integration by Parts
Let be differentiable on the closed interval . Then

$\displaystyle \int_{a}^{b} f(x)g'(x) = \left[f(x)g(x)\right]_{a}^{b} - \int_{a}^{b} f'(x)g(x)dx$

Example 3..16 Evaluate the following integrals.
1. $\displaystyle{\int_{0}^{1}3x^{2}(x^{3}+1)^{4}dx}$ 2. $\displaystyle{\int_{0}^{1}xe^{x}\ dx}$

SOLUTION 1. Let . Then . Thus we can express the integrand as . Furthermore, the limit of integration becomes $\begin{displaymath}\begin{array}{c\vert ccc} x & 0 & \to & 1\\ \hline t & 1 & \to & 2 \end{array}\end{displaymath}$ .
Thus, $\displaystyle{\int_{0}^{1}3x^{2}(x^{3}+1)^{4}dx = \int_{1}^{2}t^4 dt = \left[\f... ... \right ]_{1}^{2} = \frac{32 - 1}{5} = \frac{31}{5}}\ensuremath{\ \blacksquare}$
2. $\left\{\begin{array}{ccc} f = x && g = e^x\\ &\searrow&\\ f' = 1 &\leftarrow& g' = e^x \end{array}\right.$ Then

$\displaystyle \int_{0}^{1}xe^x dx = [xe^x]_{0}^{1} - \int_{0}^{1}e^x dx = e - [e^x]_{0}^{1} = e - (e - 1) = 1.$

Exercise 3..16 Evaluate the following definite integrals.

$\displaystyle \int_{0}^{\frac{\pi}{2}}\frac{1}{\sqrt{1+\cos{x}}}dx$

Let . Then, and . . Now need to express by . , . Thus, SOLUTION $\displaystyle{u = \sqrt{1+\cos{x}}}$ $\displaystyle{u^2 = 1+\cos{x}}$ $\displaystyle{2udu = -\sin{x}dx}$ $\displaystyle{dx = -\frac{2udu}{\sin{x}}}$ $\sin{x}$ $\displaystyle{u = \sqrt{1+\cos{x}}}$ $\displaystyle{\cos{x} = u^2 -1}$

$\displaystyle \sin{x} = \sqrt{1 - \cos^{2}{x}} = \sqrt{1- (t^2 -1)^2} = \sqrt{t^2(2 - t^2)}.$
For , . . Thus, $\sin{x} = t\sqrt{2-t^2}$ $\begin{displaymath}\begin{array}{c\vert ccc} x & 0 \to \frac{\pi}{2}\\ \hline t & \sqrt{2} \to 1 \end{array}\end{displaymath}$ $\displaystyle{dx = -\frac{2tdt}{\sin{x}} = -\frac{2tdt}{t\sqrt{2-t^2}}}$

$\displaystyle \int_{0}^{1}\frac{1}{\sqrt{1+\cos{x}}}dx = -2\int_{\sqrt{2}}^{1}\frac{1}{t\sqrt{2-t^2}}dt = 2\int_{1}^{\sqrt{2}}\frac{1}{t\sqrt{2-t^2}}dt.$
This integral is in the form , , , . Since the limit of integral is , $\displaystyle{t = \sqrt{2}\sin{\theta}}$ $\displaystyle{dt = \sqrt{2}\cos{\theta}d\theta}$ $\displaystyle{\sqrt{2-t^2} = \sqrt{2}\cos{\theta}}$ $\begin{displaymath}\begin{array}{c\vert cccc} t & 1 & \to & \sqrt{2}\\ \hline \theta & \frac{\pi}{4} & \to & \frac{\pi}{2} \end{array}\end{displaymath}$

$\displaystyle \int_{1}^{\sqrt{2}}\frac{1}{t\sqrt{2-t^2}}dt = 2\int_{\frac{\pi}{... ...ta = \sqrt{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{\sin{\theta}}d\theta.$
By trigonometric integration[1]2. multiply both numerator and denominator by , $\sin{\theta}$

$\displaystyle \sqrt{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{\sin{\theta}}d\theta$ $\displaystyle =$ $\displaystyle \sqrt{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\sin{\theta}}{\... ...int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\sin{\theta}}{1-\cos^2{\theta}}d\theta$

Now let . Then , . Thus, $u = \cos{\theta}$ $du = -\sin{\theta}d\theta$ $\begin{displaymath}\begin{array}{c\vert ccc} \theta & \frac{\pi}{4} & \to \frac{\pi}{2}\\ \hline u & \frac{\sqrt{2}}{2} & \to & 0 \end{array}\end{displaymath}$

$\displaystyle \sqrt{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\sin{\theta}}{1-\cos^2{\theta}}d\theta$ $\displaystyle =$ $\displaystyle \sqrt{2}\int_{\frac{\sqrt{2}}{2}}^{0}\frac{-1}{1-u^2}\;du = \sqrt{2}\int_{0}^{\frac{\sqrt{2}}{2}}\frac{1}{1-u^2}\; du$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{2}\left[\log\vert\frac{1+u}{1-u}\vert\right]_{0}^... ...1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}\vert\ensuremath{\ \blacksquare}$

Alternative Solution $1 + \cos{x} = 2\sin^{2}{\frac{x}{2}}$

$\displaystyle \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1 + \cos{x}}}\:dx$ $\displaystyle =$ $\displaystyle \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{2}\cos{\frac{x}{2}}}\:dx = \... ...rt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos{\frac{x}{2}}}{\cos^{2}{\frac{x}{2}}}\:dx$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos{\frac{x}{2}}}{... ...rac{x}{2}}}\:dx = \frac{1}{\sqrt{2}}\int_0^{\frac{\sqrt{2}}{2}}\frac{dt}{1-t^2}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2}}{2}\left[\log\vert\frac{1+t}{1-t}\vert\right]_0^{\... ...1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}\vert\ensuremath{\ \blacksquare}$

Properties of Definite Integral Suppose that is continuous on the limit of integration. 1. If is even function, then
$\displaystyle{\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx}$
2. If is odd function, then
$\displaystyle{\int_{-a}^{a}f(x)dx = 0}$
3.
$\displaystyle{\int_{0}^{\frac{\pi}{2}}f(\cos{x})dx = \int_{0}^{\frac{\pi}{2}}f(\sin{x})dx}$
where
$4. \ \displaystyle{I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx = \int_{0}^{\f... ...2} & n\ \mbox{even} \\ \frac{(n-1)!!}{n!!} & n\ \mbox{odd} \end{array}\right.}$
$\displaystyle{n!! = \left\{\begin{array}{ll} n\cdot(n-2)\cdot(n-4)\cdots4\cdot2... ...even}\\ n\cdot(n-2)\cdot(n-4)\cdots3\cdot1 & n\ \mbox{odd} \end{array}\right.}$

Example ..317 Show the properties of definite integral1,3,4 .
1. . Now is even function and . Thus, SOLUTION $\int_{-a}^{a}f(x)dx = \int_{-a}^{0} f(x) dx + \int_{0}^{a}f(x) dx$

$\displaystyle \int_{-a}^{a}f(x)dx = \int_{-a}^{0}f(-x)dx + \int_{0}^{a}f(x) dx.$
Here let . Then . . . Thus, $\begin{displaymath}\begin{array}{c\vert ccc} x & -a &\to & 0\\ \hline t & a &\to &0 \end{array}\end{displaymath}$ $\int_{-a}^{0}f(-x)dx = -\int_{a}^{0}f(t)dt = \int_{0}^{a}f(t) dt$

$\displaystyle \int_{-a}^{a}f(x)dx = 2 \int_{0}^{a}f(x) dx\ensuremath{\ \blacksquare}$
3. Let . Then , , $t = \cos{x}$ $dt = -\sin{x}dx = -\sqrt{1 - t^2}dx$ $\begin{displaymath}\begin{array}{c\vert ccc} x & 0 & \to & \frac{\pi}{2}\\ \hline t & 1 & \to & 0 \end{array}\end{displaymath}$

$\displaystyle \int_0^\frac{\pi}{2}f(\cos{x})dx = \int_1^0 f(t)(-\frac{1}{\sqrt{1-t^2}})dt = \int_0^1 \frac{f(t)}{\sqrt{1-t^2}}dt .$
Note the insdie of the square root is the difference of two squares. Thus and . Also, , , . Thus, $t = \sin{u},\ (0 \leq t \leq 1)$ $dt = \cos{u}du$ $u = \sin^{-1}{t}$ $du = \frac{1}{\sqrt{1-t^2}}\:dt$ $\begin{displaymath}\begin{array}{c\vert ccc} t & 0 & \to & 1\\ \hline u & 0 & \to & \frac{\pi}{2} \end{array}\end{displaymath}$

$\displaystyle \int_0^\frac{\pi}{2}f(\cos{x})dx = \int_0^1 \frac{f(t)}{\sqrt{1-t^2}}dt = \int_0^\frac{\pi}{2}f(\sin{u})du.$
Finally, $\int_0^\frac{\pi}{2}f(\sin{u})du = \int_0^\frac{\pi}{2}f(\sin{x})dx$

$\displaystyle \int_0^\frac{\pi}{2}f(\cos{x})dx = \int_0^\frac{\pi}{2}f(\sin{x})dx\ensuremath{\ \blacksquare}$

4. By 3. . We show . For , $\int_{0}^{\pi/2}\sin^{n}{x}dx = \int_{0}^{\pi/2}\cos^{n}{x}dx$ $\int_{0}^{\pi/2}\sin^{n}{x}dx$
$n \geq 2$

$\displaystyle I_{n}$ $\displaystyle =$ $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{n-1}{x}\sin{x}dx = -\left[\sin^{n-1}{x}\cos{x}\right]_{0}^{\frac{\pi}{2}}$ $\displaystyle +$ $\displaystyle (n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}\cos^{2}{x}dx$
Note that . Thus . Now we take care of the rest. $\sin{0} = 0, \cos(\frac{\pi}{2}) = 0$ $\left[\sin^{n-1}{x}\cos{x}\right]_{0}^{\frac{\pi}{2}} = 0$

$\displaystyle (n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}\cos^{2}{x}dx$ $\displaystyle =$ $\displaystyle (n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}(1-\sin^{2}{x})dx$ $\displaystyle =$ $\displaystyle (n-1)(I_{n-2} - I_{n}).$
Then we have the recurrence ralation . Note , $\displaystyle{I_{n} = \frac{n-1}{n}I_{n-2}}$ $I_{0} = \int_{0}^{\frac{\pi}{2}}dx = \frac{\pi}{2}, \ \ I_{1} = \int_{0}^{\frac{\pi}{2}}\sin{x}dx = 1$

$\displaystyle I_{n}$ $\displaystyle =$ $\displaystyle \frac{I_{n}}{I_{n-2}}\frac{I_{n-2}}{I_{n-4}}\cdots\frac{I_{2}}{I_{0}}I_{0}$ $\displaystyle =$ $\displaystyle \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{1}{2} \frac{\pi}{2} = \frac{(n-1)!!}{n!!}\frac{\pi}{2},\ n {\rm even}$ $\displaystyle I_{n}$ $\displaystyle =$ $\displaystyle \frac{I_{n}}{I_{n-2}}\frac{I_{n-2}}{I_{n-4}}\cdots\frac{I_{3}}{I_{1}}I_{1}$ $\displaystyle =$ $\displaystyle \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{2}{3} = \frac{(n-1)!!}{n!!}, \ n {\rm odd}\ \ensuremath{\ \blacksquare}$

Exercise ..317 Evaluate the following integral.

$\displaystyle \int_{0}^{2}\sqrt{4-x^2}\ dx$
The inside of square root is of the form . Let and , , . Thus SOLUTION $x = 2\sin{t}$ $t = \sin^{-1}{\frac{x}{2}}$ $dx = 2\cos{t}dt, x^2 = 4\sin^2{t}, \sqrt{4-x^2} = 2\cos{t}$ $\begin{displaymath}\begin{array}{c\vert ccc} x & 0 \to & 2\\ \hline t & 0 \to & \frac{\pi}{2} \end{array}\end{displaymath}$

$\displaystyle \int_{0}^{2}\sqrt{4-x^2}\ dx$ $\displaystyle =$ $\displaystyle \int_0^\frac{\pi}{2}2\cos{t}(2\cos{t})dt = 4\int_0^\frac{\pi}{2}\cos^2{t}dt$ $\displaystyle =$ $\displaystyle 4 \frac{1!!}{2!!}\frac{\pi}{2} = \pi\ensuremath{\ \blacksquare}$

Exercise A

1.
Evaluate the following integrals?D

(a) $\displaystyle{\int_{-2}^{2}\sin{x}\ dx}$ (b) $\displaystyle{\int_{-1}^{1} \sin^{3}{x}\ dx}$ (c) $\displaystyle{\int_{0}^{\frac{\pi}{2}}\sin^3{x}dx}$ (d) $\displaystyle{\int_{0}^{\pi}\cos{2x} dx}$

(e) $\displaystyle{\int_{-1}^{1}\sqrt{x+1}dx}$

Exercise B

1.
Evaluate the following integrals?D

(a) $\displaystyle{\int_{0}^{\frac{\pi}{2}}\cos^{4}{x}\sin{x}dx}$ (b) $\displaystyle{\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{\sqrt{1+\cos{x}}} dx}$ (c) $\displaystyle{\int_{0}^{\frac{\pi}{2}}\sin^{4}{x}dx}$
(n ????)
(d) $\displaystyle{\int_{-1}^{1}x^2 \cos{x} dx}$ (e) $\displaystyle{\int_{0}^{\pi}\cos{nx}dx}$ (f) $\displaystyle{\int_{0}^{1}\frac{x^{2}}{\sqrt{4-x^{2}}}\;dx}$

(g) $\displaystyle{\int_{0}^{1}xe^{x}\ dx}$

2.
Show that ?D
$\displaystyle{\int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx = \int_{0}^{\frac{\pi}{2}}\cos^{n}{x}dx}$

3.
Let be a continuous function on the interval . Then answer the following question concerning the function ?D
$(-\infty,\infty)$ $\displaystyle{F(x) = \int_{-x}^{x}f(t)dt}$
Show that is an odd function?D
(a)
Show that is even function implies that is an odd function.
(b)
Show that implies that ?D
(c) $f(x) = \int_{-x}^{x}f(t)dt$
Show that can be represented by a sum and a difference of functions?D
(d)

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