Trigonometric Integrals
In this topic, we will study how to integrate certain combinations involving products and powers of trigonometric functions.
We consider 8 cases.
1. Integrals of the form \(\int {\cos ax\cos bxdx} ,\) \(\int {\sin ax\cos bxdx} ,\) \(\int {\sin ax\sin bxdx}\)
To evaluate integrals of products of sine and cosine with different arguments, we apply the identities
\[\cos ax\cos bx = \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) + \cos \left( {ax - bx} \right)} \right];\]
\[\sin ax\cos bx = \frac{1}{2}\left[ {\sin \left( {ax + bx} \right) + \sin \left( {ax - bx} \right)} \right];\]
\[\sin ax\sin bx = - \frac{1}{2}\left[ {\cos \left( {ax + bx} \right) - \cos \left( {ax - bx} \right)} \right].\]
2. Integrals of the form \(\int {{\sin^m}x\,{\cos^n}xdx}\)
We assume here that the powers \(m\) and \(n\) are non-negative integers.
To find an integral of this form, use the following substitutions:
- If \(m\) (the power of sine) is odd, we use the \(u-\)substitution
\[u = \cos x,\;\; du = - \sin xdx\]
and the identity
\[{\sin ^2}x + {\cos ^2}x = 1\]
to express the remaining even power of sine in \(u-\)terms.
- If \(n\) (the power of cosine) is odd, we use the \(u-\)substitution
\[u = \sin x,\;\; du = \cos xdx\]
and the identity
\[{\sin ^2}x + {\cos ^2}x = 1\]
to express the remaining even power of cosine in \(u-\)terms.
- If both powers \(m\) and \(n\) are even, we reduce the powers using the half-angle formulas
\[{\sin ^2}x = \frac{{1 - \cos 2x}}{2},\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2}.\]
The integrals of type \(\int {{{\sin }^n}xdx} \) and \(\int {{{\cos }^n}xdx} \) can be evaluated by reduction formulas
\[\int {{{\sin }^n}xdx} = - \frac{{{{\sin }^{n - 1}}x\cos x}}{n} + \frac{{n - 1}}{n}\int {{{\sin }^{n - 2}}xdx} ,\]
\[\int {{{\cos }^n}xdx} = \frac{{{{\cos }^{n - 1}}x\sin x}}{n} + \frac{{n - 1}}{n}\int {{{\cos }^{n - 2}}xdx} .\]
3. Integrals of the form \(\int {{\tan^n}xdx} \)
The power of the integrand can be reduced using the trigonometric identity
\[1 + {\tan ^2}x = {\sec ^2}x\]
and the reduction formula
\[\int {{\tan^n}xdx} = \int {{{\tan }^{n - 2}}x\,{{\tan }^2}xdx} = \int {{{\tan }^{n - 2}}x\left( {{{\sec }^2}x - 1} \right)dx} = \frac{{{{\tan }^{n - 1}}x}}{{n - 1}} - \int {{{\tan }^{n - 2}}xdx} .\]
4. Integrals of the form \(\int {{\cot^n}xdx} \)
The power of the integrand can be reduced using the trigonometric identity
\[1 + {\cot^2}x = {\csc^2}x\]
and the reduction formula
\[\int {{{\cot }^n}xdx} = \int {{{\cot }^{n - 2}}x\,{{\cot }^2}xdx} = \int {{{\cot }^{n - 2}}x\left( {{{\csc }^2}x - 1} \right)dx} = - \frac{{{{\cot }^{n - 1}}x}}{{n - 1}} - \int {{{\cot }^{n - 2}}xdx} .\]
5. Integrals of the form \(\int {{\sec^n}xdx}\)
This type of integrals can be simplified with help of the reduction formula:
\[\int {{\sec^n}xdx} = \frac{{{{\sec }^{n - 2}}x\tan x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\sec^{n - 2}}xdx} .\]
6. Integrals of the form \(\int {{\csc^n}xdx}\)
Similarly to the previous examples, this type of integrals can be simplified by the formula
\[\int {{\csc^n}xdx} = - \frac{{{\csc^{n - 2}}x \cot x}}{{n - 1}} + \frac{{n - 2}}{{n - 1}}\int {{\csc^{n - 2}}xdx} .\]
7. Integrals of the form \(\int {{\tan^m}x\,{\sec^n}xdx} \)
- If the power of the secant \(n\) is even, then using the identity
\[1 + {\tan ^2}x = {\sec ^2}x\]
the secant function is expressed as the tangent function. The factor \({\sec ^2}x\) is separated and used for transformation of the differential. As a result, the entire integral (including differential) is expressed in terms of the function \(\tan x.\)
- If both the powers \(n\) and \(m\) are odd, then the factor \(\sec x \tan x,\) which is necessary to transform the differential, is separated. Then the entire integral is expressed in terms of \(\sec x.\)
- If the power of the secant \(n\) is odd, and the power of the tangent \(m\) is even, then the tangent is expressed in terms of the secant using the identity
\[1 + {\tan ^2}x = {\sec ^2}x.\]
After this substitution, you can calculate the integrals of the secant.
8. Integrals of the form \(\int {{\cot^m}x\,{\csc^n}xdx} \)
- If the power of the cosecant \(n\) is even, then using the identity
\[1 + {\cot^2}x = {\csc ^2}x\]
the cosecant function is expressed as the cotangent function. The factor \({\csc^2}x\) is separated and used for transformation of the differential. As a result, the integrand and differential are expressed in terms of \(\cot x.\)
- If both the powers \(n\) and \(m\) are odd, then the factor \(\cot x \csc x,\) which is necessary to transform the differential, is separated. Then the integral is expressed in terms of \(\csc x.\)
- If the power of the cosecant \(n\) is odd, and the power of the cotangent \(m\) is even, then the cotangent is expressed in terms of the cosecant using the identity
\[1 + {\cot^2}x = {\csc ^2}x.\]
After this substitution, you can find the integrals of the cosecant.
Solved Problems
Example 1.
Calculate the integral \[\int {{\sin^3}xdx}.\]
Solution.
Let \(u = \cos x,\) \(du = -\sin xdx.\) Then
\[\int {{\sin^3}xdx} = \int {{\sin^2}x\sin xdx} = \int {\left( {1 - {\cos^2}x} \right)\sin xdx} = - \int {\left( {1 - {u^2}} \right)du} = \int {\left( {{u^2} - 1} \right)du} = \frac{{{u^3}}}{3} - u + C = \frac{{{{\cos }^3}x}}{3} - \cos x + C.\]
Example 2.
Evaluate the integral \[\int {{\cos^5}xdx}.\]
Solution.
Making the substitution \(u = \sin x,\) \(du = \cos xdx\) and using the identity
\[{\cos ^2}x = 1 - {\sin ^2}x,\]
we obtain:
\[\int {{\cos^5}xdx} = \int {{{\left( {{\cos^2}x} \right)}^2}\cos xdx} = \int {{{\left( {1 - {{\sin }^2}x} \right)}^2}\cos x dx} = \int {{{\left( {1 - {u^2}} \right)}^2}du} = \int {\left( {1 - 2{u^2} + {u^4}} \right)du} = u - \frac{{2{u^3}}}{3} + \frac{{{u^5}}}{5} + C = \sin x - \frac{{2{{\sin }^3}x}}{3} + \frac{{{{\sin }^5}x}}{5} + C.\]
Example 3.
Find the integral \[\int {{\sin^6}xdx}.\]
Solution.
Using the identities
\[{\sin ^2}x = \frac{{1 - \cos 2x}}{2}\;\;\text{and}\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2},\]
we can write:
\[ I = \int {{\sin^6}xdx} = \int {{{\left( {{\sin^2}x} \right)}^3}dx} = \frac{1}{8}\int {{{\left( {1 - \cos 2x} \right)}^3}dx} = \frac{1}{8}\int {\left( {1 - 3\cos 2x + 3\,{{\cos }^2}2x - {{\cos }^3}2x} \right)dx} = \frac{x}{8} - \frac{3}{8} \cdot \frac{{\sin 2x}}{2} + \frac{3}{8}\int {{\cos^2}2xdx} - \frac{3}{8}\int {{\cos^3}2xdx}. \]
Calculate the integrals in the latter expression.
\[\int {{\cos^2}2xdx} = \int {\frac{{1 + \cos 4x}}{2}dx} = \frac{1}{2}\int {\left( {1 + \cos 4x} \right)dx} = \frac{1}{2}\left( {x + \frac{{\sin 4x}}{4}} \right) = \frac{x}{2} + \frac{{\sin 4x}}{8}.\]
To find the integral \(\int {{\cos^3}2xdx},\) we make the substitution \(u = \sin 2x,\) \(du =\) \( 2\cos 2xdx.\) Then
\[u = \sin 2x,\;du = 2\cos 2xdx.\]
Then
\[\int {{\cos^3}2xdx} = \frac{1}{2}\int {2{{\cos }^2}2x\cos 2xdx} = \frac{1}{2}\int {2\left( {1 - {{\sin }^2}2x} \right)\cos 2xdx} = \frac{1}{2}\int {\left( {1 - {u^2}} \right)du} = \frac{u}{2} - \frac{{{u^3}}}{6} = \frac{{\sin 2x}}{2} - \frac{{{{\sin }^3}2x}}{6}.\]
Hence, the initial integral is
\[ I = \frac{x}{8} - \frac{{3\sin 2x}}{{16}} + \frac{3}{8}\left( {\frac{x}{2} + \frac{{\sin 4x}}{8}} \right) - \frac{1}{8}\left( {\frac{{\sin 2x}}{2} - \frac{{{{\sin }^3}2x}}{6}} \right) + C = \frac{{5x}}{{16}} - \frac{{\sin 2x}}{4} + \frac{{3\sin 4x}}{{64}} + \frac{{{{\sin }^3}2x}}{{48}} + C. \]
Example 4.
Find the integral \[\int {{{\sin }^2}x\,{{{\cos }^3}x}dx}.\]
Solution.
The power of cosine is odd, so we make the substitution
\[u = \sin x,\;\;du = \cos xdx.\]
We rewrite the integral in terms of \(\sin x\) to obtain:
\[\int {{{\sin }^2}x\,{{\cos }^3}xdx} = \int {{{\sin }^2}x\,{{{\cos }^2}x}\cos xdx} = \int {{{\sin }^2}x{\left( {1 - {{\sin }^2}x} \right)}\cos xdx} = \int {{u^2}\left( {1 - {u^2}} \right)du} = \int {\left( {{u^2} - {u^4}} \right)du} = \frac{{{u^3}}}{3} - \frac{{{u^5}}}{5} + C = \frac{{{{\sin }^3}x}}{3} - \frac{{{{\sin }^5}x}}{5} + C.\]
Example 5.
Calculate the integral \[\int {{{\sin }^2}x\,{{\cos }^4}xdx}.\]
Solution.
We can write:
\[I = \int {{{\sin }^2}x\,{{\cos }^4}xdx} = \int {{{\left( {\sin x\cos x} \right)}^2}{{\cos }^2}xdx} .\]
We convert the integrand using the identities
\[\sin x\cos x = \frac{{\sin 2x}}{2},\;\;\;{\cos ^2}x = \frac{{1 + \cos 2x}}{2},\;\;\;{\sin ^2}x = \frac{{1 - \cos 2x}}{2}.\]
This yields
\[I = \int {{{\left( {\frac{{\sin 2x}}{2}} \right)}^2}\frac{{1 + \cos 2x}}{2}dx} = \frac{1}{8}\int {{{\sin }^2}2x\left( {1 + \cos 2x} \right)dx} = \frac{1}{8}\int {{{\sin }^2}2xdx} + \frac{1}{8}\int {{{\sin }^2}2x\cos 2xdx} = \frac{1}{8}\int {\frac{{1 - \cos 4x}}{2}dx} + \frac{1}{{16}}\int {2{{\sin }^2}2x\cos 2xdx} = \frac{1}{{16}}\int {\left( {1 - \cos 4x} \right)dx} + \frac{1}{{16}}\int {{{\sin }^2}2x\,d\left( {\sin 2x} \right)} = \frac{1}{{16}}\left( {x - \frac{{\sin 4x}}{4}} \right) + \frac{1}{{16}} \cdot \frac{{{{\sin }^3}2x}}{3} + C = \frac{x}{{16}} - \frac{{\sin 4x}}{{64}} + \frac{{{{\sin }^3}2x}}{{48}} + C.\]
Example 6.
Evaluate the integral \[\int {{{\sin }^3}x\,{{\cos }^4}xdx}.\]
Solution.
As the power of sine is odd, we use the substitution
\[u = \cos x,\;\;du = - \sin xdx.\]
The integral is written as
\[I = \int {{{\sin }^3}x\,{{\cos }^4}xdx} = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} .\]
By the Pythagorean identity,
\[{\sin ^2}x = 1 - {\cos ^2}x.\]
Hence
\[I = \int {{{\sin }^2}x\,{{\cos }^4}x\sin xdx} = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^4}x\sin xdx} = - \int {\left( {1 - {u^2}} \right){u^4}du} = \int {\left( {{u^6} - {u^4}} \right)du} = \frac{{{u^7}}}{7} - \frac{{{u^5}}}{5} + C = \frac{{{{\cos }^7}x}}{7} - \frac{{{{\cos }^5}x}}{5} + C.\]
Example 7.
Evaluate the integral \[\int {{{\sin }^3}x\,{{\cos }^5}xdx}.\]
Solution.
We see that both powers are odd, so we can substitute either \(u = \sin x\) or \(u = \cos x.\) Choosing the least exponent, we have
\[u = \cos x,\;\;du = - \sin xdx.\]
The integral takes the form
\[I = \int {{{\sin }^3}x\,{{\cos }^5}xdx} = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} .\]
Using the Pythagorean identity
\[{\sin ^2}x = 1 - {\cos ^2}x,\]
we can write
\[I = \int {{{\sin }^2}x\,{{\cos }^5}x\sin xdx} = \int {\left( {1 - {{\cos }^2}x} \right){{\cos }^5}x\sin xdx} = - \int {\left( {1 - {u^2}} \right){u^5}du} = \int {\left( {{u^7} - {u^5}} \right)du} = \frac{{{u^8}}}{8} - \frac{{{u^6}}}{6} + C = \frac{{{{\cos }^8}x}}{8} - \frac{{{{\cos }^6}x}}{6} + C.\]
Example 8.
Evaluate the integral \[\int {{{\sin }^3}x\,{{\cos }^3}xdx}.\]
Solution.
The powers of both sine and cosine are odd. Hence we can use the substitution \(u = \sin x\) or \(u = \cos x.\) Let's apply the substitution \(u = \sin x.\) Then \(du = \cos x dx,\) and the integral becomes
\[I = \int {{{\sin }^3}x\,{{\cos }^3}xdx} = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} .\]
By the Pythagorean identity,
\[{\sin ^2}x = 1 - {\cos ^2}x,\]
so we obtain
\[I = \int {{{\sin }^3}x\,{{\cos }^2}x\cos xdx} = \int {{{\sin }^3}x\left( {1 - {{\sin }^2}x} \right)\cos xdx} = \int {{u^3}\left( {1 - {u^2}} \right)du} = \int {\left( {{u^3} - {u^5}} \right)du} = \frac{{{u^4}}}{4} - \frac{{{u^6}}}{6} + C = \frac{{{{\sin }^4}x}}{4} - \frac{{{{\sin }^6}x}}{6} + C.\]
See more problems on Page 2.
