Trigonometric Integrals
Solved Problems
Example 9.
Find the integral \[\int {{{\sin }^3}x\sqrt {\cos x} dx}.\]
Solution.
Making the substitution \(u = \cos x,\) \(du = -\sin xdx\) and expressing the sine in terms of cosine with help of the formula
\[{\sin ^2}x = 1 - {\cos ^2}x,\]
we obtain
\[\int {{{\sin }^3}x\sqrt {\cos x} dx} = \int {{{\sin }^2}x\sqrt {\cos x} \sin xdx} = \int {\left( {1 - {\cos^2}x} \right)} {\sqrt {\cos x} \sin xdx} = - \int {\left( {1 - {u^2}} \right)\sqrt u du} = - \int {\left( {{u^{\frac{1}{2}} - u^{\frac{5}{2}}}} \right)du} = \frac{{{u^{\frac{5}{2} + 1}}}}{{\frac{5}{2} + 1}} - \frac{{{u^{\frac{1}{2} + 1}}}}{{\frac{1}{2} + 1}} + C = \frac{2}{7}{u^{\frac{7}{2}}} - \frac{2}{3}{u^{\frac{3}{2}}} + C = \frac{2}{7}\sqrt {{{\cos }^7}x} - \frac{2}{3}\sqrt {{{\cos }^3}x} + C.\]
Example 10.
Evaluate the integral \[\int {\sin 2x\cos 5xdx}.\]
Solution.
We use the trig identity
\[\sin ax\cos bx = \frac{1}{2}\left[ {\sin \left( {ax + bx} \right) + \sin \left( {ax - bx} \right)} \right].\]
Then the integrand is written in the form
\[\sin 2x\cos 5x = \frac{1}{2}\left[ {\sin 7x + \sin \left( { - 3x} \right)} \right] = \frac{1}{2}\left[ {\sin 7x - \sin 3x} \right].\]
Plug this expression in the integral and evaluate:
\[\int {\sin 2x\cos 5xdx} = \frac{1}{2}\int {\left( {\sin 7x - \sin 3x} \right)dx} = \frac{1}{2}\left[ {\left( { - \frac{{\cos 7x}}{7}} \right) - \left( { - \frac{{\cos 3x}}{3}} \right)} \right] + C = \frac{{\cos 3x}}{6} - \frac{{\cos 7x}}{{14}} + C.\]
Example 11.
Evaluate the integral \[\int {\sin {\frac{x}{4}} \cos {\frac{x}{3}} dx}.\]
Solution.
Convert the integrand by the formula
\[\sin ax\cos bx = \frac{1}{2}\left[ {\sin \left( {ax + bx} \right) + \sin \left( {ax - bx} \right)} \right].\]
Hence,
\[\sin \frac{x}{4}\cos \frac{x}{3} = \frac{1}{2}\left[ {\sin \left( {\frac{x}{4} + \frac{x}{3}} \right) + \sin \left( {\frac{x}{4} - \frac{x}{3}} \right)} \right] = \frac{1}{2}\left( {\sin \frac{{7x}}{{12}} - \sin \frac{x}{{12}}} \right).\]
Then the integral becomes
\[\int {\sin \frac{x}{4}\cos \frac{x}{3}dx} = \frac{1}{2}\int {\left( {\sin \frac{{7x}}{{12}} - \sin \frac{x}{{12}}} \right)dx} = -\frac{1}{2}\left( {\frac{{\cos \frac{{7x}}{{12}}}}{{\frac{7}{{12}}}} - \frac{{\cos \frac{x}{{12}}}}{{\frac{1}{{12}}}}} \right) + C = 6\cos \frac{x}{{12}} - \frac{6}{7}\cos \frac{{7x}}{{12}} + C.\]
Example 12.
Evaluate the integral \[\int {\cos \frac{x}{2} \cos \frac{x}{3} dx}.\]
Solution.
We convert the integrand into the sum using the trig identity
\[\cos ax\cos bx = \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) + \cos \left( {ax - bx} \right)} \right].\]
Then the integrand is written as
\[\cos \frac{x}{2}\cos \frac{x}{3} = \frac{1}{2}\left[ {\cos \left( {\frac{x}{2} + \frac{x}{3}} \right) + \cos \left( {\frac{x}{2} - \frac{x}{3}} \right)} \right] = \frac{1}{2}\left[ {\cos \frac{{5x}}{6} + \cos \frac{x}{6}} \right].\]
Integrating yields:
\[\int {\cos \frac{x}{2}\cos \frac{x}{3}dx} = \frac{1}{2}\int {\left( {\cos \frac{{5x}}{6} + \cos \frac{x}{6}} \right)dx} = \frac{1}{2}\left[ {\frac{{\sin \frac{{5x}}{6}}}{{\frac{5}{6}}} + \frac{{\sin \frac{x}{6}}}{{\frac{1}{6}}}} \right] + C = \frac{3}{5}\sin \frac{{5x}}{6} + 3\sin \frac{x}{6} + C.\]
Example 13.
Evaluate the integral \[\int {\sin \frac{x}{3} \sin \frac{x}{4} dx}.\]
Solution.
Using the product-to-sum identity
\[\sin ax\sin bx = - \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) - \cos \left( {ax - bx} \right)} \right],\]
we rewrite the integrand in the form
\[\sin \frac{x}{3}\sin \frac{x}{4} = - \frac{1}{2}\left[ {\cos \frac{{7x}}{{12}} - \cos \frac{x}{{12}}} \right].\]
Integrating yields:
\[\int {\sin \frac{x}{3}\sin \frac{x}{4}dx} = - \frac{1}{2}\int {\left( {\cos \frac{{7x}}{{12}} - \cos \frac{x}{{12}}} \right)dx} = \frac{1}{2}\int {\left( {\cos \frac{x}{{12}} - \cos \frac{{7x}}{{12}}} \right)dx} = \frac{1}{2}\left[ {\frac{{\sin \frac{x}{{12}}}}{{\frac{1}{{12}}}} - \frac{{\sin \frac{{7x}}{{12}}}}{{\frac{7}{{12}}}}} \right] + C = 6\sin \frac{x}{{12}} - \frac{6}{7}\sin \frac{{7x}}{{12}} + C.\]
Example 14.
Evaluate the integral \[\int {{{\tan }^4}xdx}.\]
Solution.
We use the identity
\[1 + {\tan ^2}x = {\sec ^2}x\]
to transform the integral. This yields
\[\int {{{\tan }^4}xdx} = \int {{{\tan }^2}x\,{{\tan }^2}xdx} = \int {{{\tan }^2}x\left( {{{\sec }^2}x - 1} \right)dx} = \int {{{\tan }^2}x\,{{\sec }^2}xdx} - \int {{{\tan }^2}xdx} = \int {{{\tan }^2}x\,d\left( {\tan x} \right)} - \int {\left( {{{\sec }^2}x - 1} \right)dx} = \frac{{{{\tan }^3}x}}{3} - \tan x + x + C.\]
Example 15.
Find the integral \[\int {{{\cot }^4}xdx}.\]
Solution.
Using the reduction formula
\[\int {{{\cot }^n}xdx} = - \frac{{{{\cot }^{n - 1}}x}}{{n - 1}} - \int {{{\cot }^{n - 2}}xdx},\]
we can write
\[I = \int {{{\cot }^4}xdx} = - \frac{{{{\cot }^3}x}}{3} - \int {{{\cot }^2}xdx} .\]
Applying the reduction formula repeatedly yields:
\[I = - \frac{{{{\cot }^3}x}}{3} - \int {{{\cot }^2}xdx} = - \frac{{{{\cot }^3}x}}{3} - \left[ { - \cot x - \int {1dx} } \right] = - \frac{{{{\cot }^3}x}}{3} + \cot x + x + C.\]
Example 16.
Find the integral \[\int {{{\tan }^5}xdx}.\]
Solution.
To evaluate the integral we use the reduction formula
\[\int {{{\tan }^n}xdx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}xdx} .\]
Then we can write
\[I = \int {{{\tan }^5}xdx} = \frac{{{{\tan }^4}x}}{4} - \int {{{\tan }^3}xdx} .\]
Applying this formula repeatedly, we obtain
\[I = \frac{{{{\tan }^4}x}}{4} - \int {{{\tan }^3}xdx} = \frac{{{{\tan }^4}x}}{4} - \left[ {\frac{{{{\tan }^2}x}}{2} - \int {\tan xdx} } \right] = \frac{{{{\tan }^4}x}}{4} - \frac{{{{\tan }^2}x}}{2} + \int {\tan xdx} .\]
The integral of tangent is a standard integral:
\[\int {\tan xdx} = - \ln \left| {\cos x} \right|.\]
Hence
\[I = \frac{{{{\tan }^4}x}}{4} - \frac{{{{\tan }^2}x}}{2} + \int {\tan xdx} = \frac{{{{\tan }^4}x}}{4} - \frac{{{{\tan }^2}x}}{2} - \ln \left| {\cos x} \right| + C.\]
Example 17.
Calculate the integral \[\int {{{\cot }^5}xdx}.\]
Solution.
Using the identity
\[1 + {\cot ^2}x = {\csc ^2}x,\]
we have
\[\int {{\cot^5}xdx} = \int {{\cot^3}x\,{{\cot }^2}xdx} = \int {{\cot^3}x\left( {{\csc^2}x - 1} \right)dx} = \int {{\cot^3}x\,{{\csc }^2}xdx} - \int {{\cot^3}xdx} = - \int {{\cot^3}x\,d\left( {\cot x} \right)} - \int {\cot x \,{\cot^2}xdx} = - \frac{{{\cot^4}x}}{4} - \int {\cot x\left( {{\csc^2}x - 1} \right)dx} = - \frac{{{\cot^4}x}}{4} - \int {\cot x \,{\csc^2}xdx} + \int {\cot xdx} = - \frac{{{\cot^4}x}}{4} - \int {\cot x \,d\left( {\cot x} \right)} + \int {\frac{{d\left( {\sin x} \right)}}{{\sin x}}} = - \frac{{{\cot^4}x}}{4} + \frac{{{\cot^2}x}}{2} + \ln \left| {\sin x} \right| + C.\]
Example 18.
Calculate the integral \[\int {{{\sec }^3}xdx}.\]
Solution.
We use the reduction formula
\[
\int {{\sec^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\sec^{n - 2}}xdx}.\]
Hence,
\[\int {{\sec^3}xdx} = \frac{{\sec x\tan x}}{2} + \frac{1}{2}\int {\sec xdx}.\]
The integral \(\int {\sec xdx}\) is a table integral which is equal to
\[\int {\sec xdx} = \ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right|.\]
(It can be easily found using the universal trigonometric substitution \(\tan {\frac{x}{2}}\).) As a result, the initial integral becomes
\[\int {{\sec^3}xdx} = \frac{{\sec x\tan x}}{2} + \frac{1}{2}\ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right| + C.\]
Example 19.
Find the integral \[\int {{{\csc }^3}xdx}.\]
Solution.
We use the reduction formula
\[\int {{{\csc }^n}xdx} = - \frac{{{{\csc }^{n - 2}}x\cot x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\csc }^{n - 2}}xdx} .\]
Then
\[I = \int {{{\csc }^3}xdx} = - \frac{{\csc x\cot x}}{2} + \frac{1}{2}\int {\csc xdx} .\]
The integral \(\int {\csc xdx} \) is a table integral which is equal to \(\ln \left| {\tan \frac{x}{2}} \right|.\) Hence
\[I = - \frac{{\csc x\cot x}}{2} + \frac{1}{2}\int {\csc xdx} = - \frac{{\csc x\cot x}}{2} + \frac{1}{2}\ln \left| {\tan \frac{x}{2}} \right| + C.\]
Example 20.
Evaluate the integral \[\int {{{\csc }^4}xdx}.\]
Solution.
We use the reduction formula
\[\int {{\csc^n}xdx} = - \frac{{{\csc^{n - 2}}x \cot x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\csc^{n - 2}}xdx}.\]
Hence,
\[\int {{\csc^4}xdx} = - \frac{{{\csc^2}x \cot x}}{3} + \frac{2}{3}\int {{\csc^2}xdx} = - \frac{{{\csc^2}x \cot x}}{3} - \frac{2}{3}\cot x + C = - \frac{{\cot x}}{3}\left( {{{\csc }^2}x + 2} \right) + C.\]
Example 21.
Find the integral \[\int {{{\sec }^4}xdx}.\]
Solution.
We apply the reduction formula
\[\int {{{\sec }^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{{\sec }^{n - 2}}xdx} .\]
This yields:
\[I = \int {{{\sec }^4}xdx} = \frac{{{{\sec }^2}x\tan x}}{3} + \frac{2}{3}\int {{{\sec }^2}xdx} .\]
The integral \(\int {{{\sec }^2}xdx} \) is equal to \(\tan x.\) Then
\[I = \frac{{{{\sec }^2}x\tan x}}{3} + \frac{2}{3}\int {{{\sec }^2}xdx} = \frac{{{{\sec }^2}x\tan x}}{3} + \frac{2}{3}\tan x + C = \frac{{\tan x}}{3}\left( {{{\sec }^2}x + 2} \right) + C.\]
Example 22.
Compute \[\int {{\tan^3}x\,{{\sec }^2}xdx}.\]
Solution.
\[\int {{\tan^3}x\,{{\sec }^2}xdx} = \int {{\tan^3}x\,d\left( {\tan x} \right)} = \frac{{{{\tan }^4}x}}{4} + C.\]
Example 23.
Compute \[\int {{\tan^2}x\sec xdx}.\]
Solution.
We use the identity
\[1 + {\tan ^2}x = {\sec ^2}x.\]
Then
\[I = \int {{\tan^2}x\sec xdx} = \int {\left( {{\sec^2}x - 1} \right)\sec xdx} = \int {{{\sec }^3}xdx} - \int {\sec xdx} .\]
Since (see Example \(18\))
\[\int {{\sec^3}xdx} = \frac{{\sec x\tan x}}{2} + \frac{1}{2} \ln {\left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right|}\]
and \(\int {\sec xdx}\) is a table integral equal to
\[\int {\sec xdx} = \ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right|,\]
we obtain the following answer:
\[I = \int {{{\sec }^3}xdx} - \int {\sec xdx} = \frac{{\sec x\tan x}}{2} + \frac{1}{2}\ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right| - \ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right| + C = \frac{{\sec x\tan x}}{2} - \frac{1}{2}\ln \left| {\tan \left( {\frac{x}{2} + \frac{\pi }{4}} \right)} \right| + C.\]
