Integration of Trigonometric Functions

Integration of Trigonometric Functions[I]

$\displaystyle \int \sin^{m}{x}\cos^{n}{x}\: dx.$
1. For . Let $t = \cos{x}$ . Then $dt = -\sin{x}$ and

$\displaystyle \int \sin^{m}{x}\cos^{n}{x}\: dx = \int \cos^{n}{x}\sin{x}dx = \int t^n (-dt) = -\int t^n \:dt$
For . Let $t = \sin{x}$ . Then $dt = \cos{x}$ and

$\displaystyle \int \sin^{m}{x}\cos^{n}{x}\: dx = \int \sin^{m}{x}\cos{x}dx = \int t^m dt.$

Example 3..8 Integrate the following function.

$\displaystyle \sin^{3}{x}\cos{x}$
.
SOLUTION Let $t = \sin{x}$ . Thne $dt = \cos{x}dx$ ,

$\displaystyle \int \sin^{3}{x}\cos{x}\: dx = \int t^3 dt = \frac{t^4}{4} + c = \frac{\sin^{4}{x}}{4} + c\ensuremath{\ \blacksquare}$

Exercise 3..8 Integrate the following function.

$\displaystyle \sin{x}\cos^{4}{x}$
SOLUTION Let $t = \cos{x}$ . Then $dt = -\sin{x}dx$ .

$\displaystyle \int \sin{x}\cos^{4}{x}\: dx = \int t^4 (-dt) = -\frac{t^5}{5} + c = -\frac{\cos^{5}{x}}{5} + c\ensuremath{\ \blacksquare}$

Integration of Trigonometric Functions[I]

$\displaystyle \int \sin^{m}{x}\cos^{n}{x}\: dx.$
2. If is odd, then is even. Using $\sin^{2}{x} = 1 - \cos^{2}{x}$ , express $\sin^{m-1}{x}$ as in the form of $\cos{x}$ . Thus,

$\displaystyle \int \sin^{m-1}{x}\cos^{n}{x}\sin{x}\:dx = \int (1 - \cos^{2}{x})^{\frac{m-1}{2}}\cos^{n}{x}\sin{x}dx.$
Now let $t = \cos{x}$ . Then $dt = -\sin{x}dx$ and

$\displaystyle \int (1 - \cos^{2}{x})^{\frac{m-1}{2}}\cos^{n}{x}\sin{x}dx$ $\displaystyle =$ $\displaystyle \int (1 - t^2)^{\frac{m-1}{2}}t^n (-dt)$

Similarly for odd.

Example 3..9 Integrate the following function.

$\displaystyle \sin^{3}{x}\cos^{2}{x}$
Since is odd power of , SOLUTION $\sin^{3}{x}$ $\sin{x}$

$\displaystyle \int \sin^{3}{x}\cos^{2}{x}\:dx = \int (1 - \cos^{2}{x})\cos^{2}{x}\sin{x}\:dx$
Now let . Then and $t = \cos{x}$ $dt = -\sin{x}dx$

$\displaystyle \int (1 - \cos^{2}{x})\cos^{2}{x}\sin{x}\:dx$ $\displaystyle =$ $\displaystyle \int (1 - t^2)t^2 (-dt) = -\int (t^2 - t^4)dt$

$\displaystyle =$ $\displaystyle -\left(\frac{t^{3}}{3} - \frac{t^{5}}{5}\right) + c = -\frac{\cos^{3}{x}}{3} + \frac{\cos^{5}{x}}{5} + c\ensuremath{\ \blacksquare}$

Integrate the following function .
Exercise ..39 $\displaystyle{\sin^{3}{x}\cos^{3}{x} }$
Let . Then and SOLUTION $t = \sin{x}$ $dt = \cos{x}dx$

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

Integration of Trigonometric Function[I] 3. Suppose that and are both even. Now let . Then we can express by using . Consider the right triangle with the adjacent of the angle is 1 and the opposite is . Then and Also, Thus

$\displaystyle \int \sin^{m}{x}\cos^{n}{x}\: dx.$
$t = \tan{x}$ $\sin^{2}{x}, \cos^{2}{x}, dx$ $t = \tan{x}$

$\displaystyle \cos^{2}{x} = (\cos{x})^2 = (\frac{1}{\sqrt{1+t^2}})^2 = \frac{1}{1+t^2},$

$\displaystyle \sin^{2}{x} = (\sin{x})^2 = (\frac{t}{\sqrt{1+t^2}})^2 = \frac{t^2}{1+t^2}.$

$\displaystyle dt = \frac{1}{\cos^{2}{x}}dx = \frac{1}{\frac{1}{t^2+1}}dx = (t^2 + 1)dx$

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

Find the following indefinite integrals. 1. 2.
Example ..310
$\displaystyle{\sin^{2}{x}\cos^{2}{x}}$ $\displaystyle{\frac{\sin^{2}{x}}{\cos^{4}{x}} }$
1. Instead of using , it is easier to use double angle formula. SOLUTION $t = \tan{x}$

$\displaystyle \sin^{2}{x} = \frac{1 - \cos{2x}}{2},\ \cos^{2}{x} = \frac{1 + \cos{2x}}{2}$

$\displaystyle \int \sin^{2}{x}\cos^{2}{x}\:dx$ $\displaystyle =$ $\displaystyle \int \frac{(1-\cos{2x})(1+\cos{2x})}{4}\:dx$ $\displaystyle =$ $\displaystyle \int \frac{1-\cos^{2}{2x}}{4}\:dx$ $\displaystyle =$ $\displaystyle \frac{1}{4}\left(\int 1 - \frac{1 + \cos{4x}}{2}\right)\:dx = \fr... ...{4}\left(x - \frac{x}{2} - \frac{\sin{4x}}{8}\right)\ensuremath{\ \blacksquare}$
2. Let . Then , , . Thus $t = \tan{x}$ $\cos^{2}{x} = \frac{1}{1+t^2}$ $\sin^{2}{x} = \frac{t^2}{1+t^2}$ $dx = \frac{1}{t^2 + 1}dt$

$\displaystyle \int \frac{\sin^{2}{x}}{\cos^{4}{x}}dx$ $\displaystyle =$ $\displaystyle \int \frac{t^2}{1+t^2}(1+t^2)^2 \frac{1}{t^2 + 1}dt = \int t^2dt = \frac{t^3}{3} + c$ $\displaystyle =$ $\displaystyle \frac{1}{3}\tan^{3}{x} + c$

Integrate the following function. 1. 2.
Exercise ..310
$\tan^{2}{x}$ $\displaystyle{\frac{1}{\cos{x}}}$
SOLUTION

1. Let and express . Then $t = \tan{x}$ $\tan{x} = \frac{\sin{x}}{\cos{x}}$

$\displaystyle dx = \frac{1}{1 + t^{2}} dt\hskip 0.5cm \cos^{2}{x} = \frac{1}{1+t^2}\hskip 0.5cm \sin^{2}{x} = \frac{t^2}{1+t^2}.$
Thus

$\displaystyle \int \tan^{2}{x} dx$ $\displaystyle =$ $\displaystyle \int \frac{\frac{t^2}{1+t^2}}{\frac{1}{1+t^2}}\cdot \frac{1}{1+t^2}dt = \int \frac{t^{2}}{1 + t^2} dt$ $\displaystyle =$ $\displaystyle \int \frac{1 + t^{2} - 1}{1+ t^{2}} dt = \int [1 - \frac{1}{1+ t^2}] dt$ $\displaystyle =$ $\displaystyle t - \tan^{-1}{t} + c = \tan{x} - x + c$
Alternative Solution

$\displaystyle \int \tan^{2}{x} dx$ $\displaystyle =$ $\displaystyle \int \frac{\sin^{2}{x}}{\cos^{2}{x}} dx = \int \frac{1 - \cos^{2}{x}}{\cos^{2}{x}} dx$ $\displaystyle =$ $\displaystyle \int (\sec^{2}{x} - 1) dx = \tan{x} - x + c \ensuremath{\ \blacksquare}$

2. . Then it is in the form of [1]-2. $\displaystyle{\frac{1}{\cos{x}} = \frac{\cos{x}}{\cos^{2}{x}} = \frac{\cos{x}}{1-\sin^{2}{x}}}$

$\displaystyle \int \frac{1}{\cos{x}}\:dx$ $\displaystyle =$ $\displaystyle \int \frac{\cos{x}}{1-\sin^{2}{x}}\:dx$
Now, and $t = \sin{x}$ $dt = \cos{x}dx$

$\displaystyle \int \frac{1}{\cos{x}}\:dx$ $\displaystyle =$ $\displaystyle \int \frac{dt}{1-t^2} = \int \frac{1}{(1+t)(1-t)}dt.$
Using partial fraction decomposition,

$\displaystyle \frac{1}{(1+t)(1-t)} = \frac{A}{1+t} + \frac{B}{1-t}.$
Clear the denominator,

$\displaystyle 1 = A(1-t) + B(1+t) = (-A+B)t + A + B.$
which implies . Then . Thus, $A = \frac{1}{2}, B = \frac{1}{2}$

$\displaystyle \int \frac{1}{\cos{x}}\:dx$ $\displaystyle =$ $\displaystyle \frac{1}{2} \int \big(\frac{1}{1+t} + \frac{1}{1-t}\big)\:dt = \frac{1}{2} \big(\log\vert 1+t\vert - \log\vert 1-t\vert\big) + c$ $\displaystyle =$ $\displaystyle \frac{1}{2} \log\vert\frac{1+t}{1-t}\vert + c = \frac{1}{2}\log\vert\frac{1+\sin{x}}{1-\sin{x}}\vert + c.$

Integration of Trigonometric Functions[II] Let . Then Now consider the right triangle with the angle ,the adjacent to the angle 1, and opposite to the angle . Thus,

$\displaystyle \int \sin^{m}{x}\cos^{n}{x}\:dx$
$x = 2\tan^{-1}{t}$

$\displaystyle dx = d(2 \tan^{-1}{t}) = \frac{2dt}{1+t^{2}}.$
$\frac{x}{2}$

$\displaystyle \cos{\frac{x}{2}} = \frac{1}{\sqrt{1+t^2}}, \sin{\frac{x}{2}} = \frac{t}{\sqrt{1+t^2}}$

$\displaystyle \sin{x}$ $\displaystyle =$ $\displaystyle \sin{2\cdot\frac{x}{2}} = 2\sin{\frac{x}{2}}\cos{\frac{x}{2}} = 2 \frac{t}{\sqrt{1 + t^2}}\frac{1}{\sqrt{1 + t^2}} = \frac{2t}{1+t^{2}}$ $\displaystyle \cos{x}$ $\displaystyle =$ $\displaystyle \cos{2\cdot \frac{x}{2}} = \cos^{2}{\frac{x}{2}} - \sin^{2}{\frac{x}{2}} = \frac{1}{1 + t^2} - \frac{t^2}{1 + t^2} = \frac{1-t^{2}}{1+t^{2}}.$

Example ..311 Find the integral . $\displaystyle{\int \frac{1}{1+\cos{x}}}$
It is not in the form [1]. Then let , . SOLUTION $t = \tan{\frac{x}{2}}$ $x = 2\tan^{-1}{t}$

$\displaystyle dx = \frac{2}{1+ t^2}dt, \ \cos{x} = \frac{1 - t^2}{1 + t^2}.$
Thus,

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

Find the integral .
Exercise ..311 $\displaystyle{\int\sec^{3}{x}}$

SOLUTION
$\left\{\begin{array}{lcl} f = \sec{x} & & g' = \sec^2{x}\\ &\searrow& \\ f' = \sec{x}\tan{x} &\leftarrow& g = \tan{x} \end{array}\right.$

$\displaystyle \int{\sec^3{x}} \; dx$ $\displaystyle =$ $\displaystyle \int \sec^2{x}\sec{x}\; dx$ $\displaystyle =$ $\displaystyle \sec{x}\tan{x} - \sec{x}\tan^2{x}\;dx$ $\displaystyle =$ $\displaystyle \sec{x}\tan{x} - \int \sec^3{x}\; dx + \int \sec{x}\; dx.$
Now let . Then $I = \int \sec^3{x}\; dx$

$\displaystyle 2I$ $\displaystyle =$ $\displaystyle \sec{x}\tan{x} + \int \sec{x}\;dx$ $\displaystyle =$ $\displaystyle \sec{x}\tan{x} + \log\vert\sec{x} + \tan{x}\vert + c.$
Thus,

$\displaystyle I = \frac{1}{2}\left(\sec{x}\tan{x} + \log\vert\sec{x} + \tan{x}\vert \right) + c\ensuremath{\ \blacksquare}$

Exercise A

1.
Work out the following integrals?D

(a) $\displaystyle{\int{\sin^3{x}\cos{x}}\ dx}$ (b) $\displaystyle{\int{\sin^2{3x}\cos{3x}}\ dx}$ (c) $\displaystyle{\int{\cos^2{x}}\ dx}$

(d) $\displaystyle{\int{\cos^{3}{x}} \ dx }$ (e) $\displaystyle{\int{\cos^{4}{x}\sin^3{x}}\ dx}$ (f) $\displaystyle{\int{\sin{2x}\cos{3x}}\ dx}$

(g) $\displaystyle{\int{\sin{2x}\sin{x}}\ dx}$ (h) $\displaystyle{\int{\cos{x}\cos{2x}}\ dx}$ (i) $\displaystyle{\int{\tan{x}\sec^2{x}}\ dx}$

(j) $\displaystyle{\int{\sec^3{x}}\ dx}$

Exercise B

1.
Work out the following integrals?D

(a) $\displaystyle{\int{\sin^3{x}}\ dx}$ (b) $\displaystyle{\int{\sin^2{3x}}\ dx}$ (c) $\displaystyle{\int{\sin^3{x}\cos^2{x}}\ dx}$

(d) $\displaystyle{\int{\cos{3x}\sin{2x}} \ dx }$ (e) $\displaystyle{\int{\sin^5{x}}\ dx}$ (f) $\displaystyle{\int{\sec^2{\pi x}}\ dx}$ (g) $\displaystyle{\int{\tan^3{x}}\ dx}$

(h) $\displaystyle{\int{\tan^2{x}\sec^2{x}}\ dx}$ (i) $\displaystyle{\int{\tan^3{x}\sec^3{x}}\ dx}$ (j) $\displaystyle{\int{\sec^5{x}}\ dx}$

(k) $\displaystyle{\int{\frac{dx}{3 - 2\cos{x}}}}$ (l) $\displaystyle{\int{\frac{\sin{x}}{2 - \sin{x}}}\ dx}$ (m) $\displaystyle{\int{\frac{1 + \sin{x}}{1 + \cos{x}}}\ dx}$

(n) $\displaystyle{\int{\frac{\sin^2{x}}{\sin^2{x} - \cos^2{x}}}\ dx}$ (o) $\displaystyle{\int{\frac{dx}{1 + \tan{x}}}}$

Next:Integration of Irrational Functions Up:Integration Previous:Integration of Rational Functions Contents Index

$\displaystyle \int (1 - \cos^{2}{x})\cos^{2}{x}\sin{x}\:dx$	$\displaystyle =$	$\displaystyle \int (1 - t^2)t^2 (-dt) = -\int (t^2 - t^4)dt$
	$\displaystyle =$	$\displaystyle -\left(\frac{t^{3}}{3} - \frac{t^{5}}{5}\right) + c = -\frac{\cos^{3}{x}}{3} + \frac{\cos^{5}{x}}{5} + c\ensuremath{\ \blacksquare}$