Evaluating Definite Integral

u-substitution

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

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

Usage If it is impossible to find the indefinite integral of $f(x)$, then the following might help.

NOTE Let $x = \phi(u)$. Then $dx = \phi'(u)du$. Now the limit of intervals must be changed from $x = \phi(a)$ to $u = a$ and $x = \phi(b)$ to $u = b$.

$$\begin{array}{l\vert lll} x & \phi(a) & \to & \phi(b)\ \hline t & a & \to & b \end{array}$$.

Integration by Parts Let $f(x),g(x)$ be differentiable on the closed interval $[a,b]$. Then

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

Usage We let $\sin{x},\cos{x},e^{\pm x}$ be $g'(x)$.

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

SOLUTION 1. Let $t = x^3 + 1$. Then $dt = 3x^2 dx$. Thus we can express the integrand as $t^4$. Furthermore, the limit of integration becomes $$\begin{array}{c\vert ccc} x & 0 & \to & 1\ \hline t & 1 & \to & 2 \end{array}$$.
Thus, $${\int_{0}^{1}3x^{2}(x^{3}+1)^{4}dx = \int_{1}^{2}t^4 dt = \left[\f... ... \right ]_{1}^{2} = \frac{32 - 1}{5} = \frac{31}{5}}\ensuremath{ \blacksquare}$$

2. $\left\{\begin{array}{ccc} f = x && g = e^x\\ &\searrow&\\ f' = 1 &\leftarrow& g' = e^x \end{array}\right.$ Then

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

If we let $u = e^x$, then $du = e^x dx$ and $x = \log{u}$. Thus we can not solve this by u-substitution directly.

Exercise 3..16   Evaluate the following definite integrals.

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

Let $u = \sqrt{1+ \cos{x}}$.

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

Since $\cos^{2}{x} + \sin^{2}{x} = 1$, $\sin^{2}{x} = 1 - \cos^{2}{x}$. Thus, $\sin{x} = \pm \sqrt{1 - \cos^{2}{x}}$. Note $0 \leq x \leq 1$. Thus $\sin{x} = \sqrt{1 - \cos^{2}{x}}$.

$$\sin{x} = \sqrt{1 - \cos^{2}{x}} = \sqrt{1- (t^2 -1)^2} = \sqrt{t^2(2 - t^2)}.$$

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

Check

$1-(t^2-1)^2 = (1+t^2-1)(1-(t^2-1)) = t^2(2-t^2)$.

$$\int_{0}^{1}\frac{1}{\sqrt{1+\cos{x}}}dx = -2\int_{\sqrt{2}}^{1}\frac{1}{t\sqrt{2-t^2}}dt = 2\int_{1}^{\sqrt{2}}\frac{1}{t\sqrt{2-t^2}}dt.$$

This integral is in the form $a^2 - x^2$, $${t = \sqrt{2}\sin{\theta}}$$, $${dt = \sqrt{2}\cos{\theta}d\theta}$$, $${\sqrt{2-t^2} = \sqrt{2}\cos{\theta}}$$.

$\sqrt{2-t^2} = \sqrt{2-(\sqrt{2}\sin{\theta})^2} = \sqrt{2(1-\sin^{2}{\theta})} = \sqrt{2}\cos{\theta}$

Figure 3.10: Check

Since the limit of integral is $$\begin{array}{c\vert cccc} t & 1 & \to & \sqrt{2}\ \hline \theta & \frac{\pi}{4} & \to & \frac{\pi}{2} \end{array}$$,

$$\int_{1}^{\sqrt{2}}\frac{1}{t\sqrt{2-t^2}}dt = 2\int_{\frac{\pi}{... ...ta = \sqrt{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{1}{\sin{\theta}}d\theta.$$

By trigonometric integration[1]2. multiply both numerator and denominator by $\sin{\theta}$ ,

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

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

$$\sqrt{2}\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}\frac{\sin{\theta}}{1-\cos^2{\theta}}d\theta = \sqrt{2}\int_{\frac{\sqrt{2}}{2}}^{0}\frac{-1}{1-u^2}\;du = \sqrt{2}\int_{0}^{\frac{\sqrt{2}}{2}}\frac{1}{1-u^2}\; du$$

$$= \frac{\sqrt{2}}{2}\left[\log\vert\frac{1+u}{1-u}\vert\right]_{0}^... ...1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}\vert\ensuremath{ \blacksquare}$$

u-substitution

In u-substitution, complexity of integration depends on the choice of $u$. Alternative Solution $1 + \cos{x} = 2\sin^{2}{\frac{x}{2}}$

$$\int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{1 + \cos{x}}}\:dx = \int_0^{\frac{\pi}{2}}\frac{1}{\sqrt{2}\cos{\frac{x}{2}}}\:dx = \... ...rt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos{\frac{x}{2}}}{\cos^{2}{\frac{x}{2}}}\:dx$$

$$= \frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos{\frac{x}{2}}}{... ...rac{x}{2}}}\:dx = \frac{1}{\sqrt{2}}\int_0^{\frac{\sqrt{2}}{2}}\frac{dt}{1-t^2}$$

$$= \frac{\sqrt{2}}{2}\left[\log\vert\frac{1+t}{1-t}\vert\right]_0^{\... ...1 + \frac{\sqrt{2}}{2}}{1 - \frac{\sqrt{2}}{2}}\vert\ensuremath{ \blacksquare}$$

Properties of Definite Integral Suppose that $f(x)$ is continuous on the limit of integration.
1. If $f(x)$ is even function, then $${\int_{-a}^{a}f(x)dx = 2\int_{0}^{a}f(x)dx}$$

2. If $f(x)$ is odd function, then $${\int_{-a}^{a}f(x)dx = 0}$$

3. $${\int_{0}^{\frac{\pi}{2}}f(\cos{x})dx = \int_{0}^{\frac{\pi}{2}}f(\sin{x})dx}$$

$4. \displaystyle{I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{n}{x}dx = \int_{0}^{\f... ...2} & n \mbox{even} \\ \frac{(n-1)!!}{n!!} & n \mbox{odd} \end{array}\right.}$
where $${n!! = \left\{\begin{array}{ll} n\cdot(n-2)\cdot(n-4)\cdots4\cdot2... ...even}\\ n\cdot(n-2)\cdot(n-4)\cdots3\cdot1 & n \mbox{odd} \end{array}\right.}$$

Even/Odd Functions $f(x)$ is even function means that $f(-x) = f(x)$ and the graph of a function $f(x)$ is symmetric with respect to the $y$-axis. Thus, $\int_{-a}^{a}f(x)dx = 2\int_0^a f(x)dx$.
$f(x)$ is odd function means that $f(-x) = -f(x)$ and the graph of a function $f(x)$ is symmetric with respect to the $x$-axis. Thus, $\int_{-a}^{a}f(x)dx = 0$.

Example 3..17   Show the properties of definite integral1,3,4 .

SOLUTION 1. $\int_{-a}^{a}f(x)dx = \int_{-a}^{0} f(x) dx + \int_{0}^{a}f(x) dx$. Now $f(x)$ is even function and $f(x) = f(-x)$. Thus,

$$\int_{-a}^{a}f(x)dx = \int_{-a}^{0}f(-x)dx + \int_{0}^{a}f(x) dx.$$

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

$$\int_{-a}^{a}f(x)dx = 2 \int_{0}^{a}f(x) dx\ensuremath{ \blacksquare}$$

3. Let $t = \cos{x}$. Then

Since $0 \leq x \leq \frac{\pi}{2}$, we have $\sin{x} \geq 0$ and $\sin{x} = \sqrt{1 - \cos^{2}{x}} = \sqrt{1 - t^2}$. $dt = -\sin{x}dx = -\sqrt{1 - t^2}dx$, $$\begin{array}{c\vert ccc} x & 0 & \to & \frac{\pi}{2}\ \hline t & 1 & \to & 0 \end{array}$$,

Check

$\int_a^b f(x)dx = -\int_b^a f(x)dx$.

$$\int_0^\frac{\pi}{2}f(\cos{x})dx = \int_1^0 f(t)(-\frac{1}{\sqrt{1-t^2}})dt = \int_0^1 \frac{f(t)}{\sqrt{1-t^2}}dt .$$

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

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

Finally, $\int_0^\frac{\pi}{2}f(\sin{u})du = \int_0^\frac{\pi}{2}f(\sin{x})dx$

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

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

Check

$f(x) = \sin^{n-1}{x}, g'(x) = \sin{x}$, $f'(x) = (n-1)\sin^{n-2}{x}\cos{x}, g(x) = -\cos{x}$.

$$I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{n-1}{x}\sin{x}dx = -\left[\sin^{n-1}{x}\cos{x}\right]_{0}^{\frac{\pi}{2}}$$

$$+ (n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}\cos^{2}{x}dx$$

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

$$(n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}\cos^{2}{x}dx = (n-1)\int_{0}^{\frac{\pi}{2}}\sin^{n-2}{x}(1-\sin^{2}{x})dx$$

$$= (n-1)(I_{n-2} - I_{n}).$$

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

Check

$I_2 = \frac{2-1}{2}I_{2-2} = \frac{1}{2}\cdot \frac{\pi}{2}$
$I_3 = \frac{3-1}{3}I_{3-2} = \frac{2}{3}\cdot 1$
$I_4 = \frac{4-1}{4}I_{4-2} = \frac{3}{4}\cdot \frac{\pi}{4}$
$I_5 = \frac{5-1}{5}I_{5-2} = \frac{4}{5}\cdot \frac{2}{3}$.

$$I_{n} = \frac{I_{n}}{I_{n-2}}\frac{I_{n-2}}{I_{n-4}}\cdots\frac{I_{2}}{I_{0}}I_{0}$$

$$= \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{1}{2} \frac{\pi}{2} = \frac{(n-1)!!}{n!!}\frac{\pi}{2}, n {\rm even}$$

$$I_{n} = \frac{I_{n}}{I_{n-2}}\frac{I_{n-2}}{I_{n-4}}\cdots\frac{I_{3}}{I_{1}}I_{1}$$

$$= \frac{n-1}{n} \frac{n-3}{n-2} \cdots \frac{2}{3} = \frac{(n-1)!!}{n!!}, n {\rm odd} \ensuremath{ \blacksquare}$$

Exercise 3..17   Evaluate the following integral.

$$\int_{0}^{2}\sqrt{4-x^2} dx$$

SOLUTION This problem can be thought of finding the area of a quarter of circle. Thus it's area is $\pi$. The inside of square root is of the form $a^2 - x^2$. Let $x = 2\sin{t}$ and $t = \sin^{-1}{\frac{x}{2}}$, $dx = 2\cos{t}dt, x^2 = 4\sin^2{t}, \sqrt{4-x^2} = 2\cos{t}$, $$\begin{array}{c\vert ccc} x & 0 \to & 2\ \hline t & 0 \to & \frac{\pi}{2} \end{array}$$. Thus

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

$$= 4 \frac{1!!}{2!!}\frac{\pi}{2} = \pi\ensuremath{ \blacksquare}$$

Exercise A

1.

Evaluate the following integrals

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

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

Exercise B

1.

Evaluate the following integrals

(a) $${\int_{0}^{\frac{\pi}{2}}\cos^{4}{x}\sin{x}dx}$$ (b) $${\int_{0}^{\frac{\pi}{2}}\frac{\sin{x}}{\sqrt{1+\cos{x}}} dx}$$ (c) $${\int_{0}^{\frac{\pi}{2}}\sin^{4}{x}dx}$$

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

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

2.

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

3.

Let $f(x)$ be a continuous function on the interval $(-\infty,\infty)$. Then answer the following question concerning the function $${F(x) = \int_{-x}^{x}f(t)dt}$$

(a) Show that $F(x)$ is an odd function

(b) Show that $f(x)$ is even function implies that $f'(x)$ is an odd function.

(c) Show that $f(x) = \int_{-x}^{x}f(t)dt$ implies that $f(x) = 0$

(d) Show that $f(x)$ can be represented by a sum and a difference of functions