Calculus/Integration techniques/Trigonometric Integrals

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Integration techniques/Trigonometric Integrals

When the integrand is primarily or exclusively based on trigonometric functions, the following techniques are useful.

Contents

Powers of Sine and CosineEdit

We will give a general method to solve generally integrands of the form \cos^m(x)\cdot\sin^n(x) . First let us work through an example.

\int\cos^3(x)\sin^2(x)dx

Notice that the integrand contains an odd power of cos. So rewrite it as

\int\cos^2(x)\sin^2(x)\cos(x)dx

We can solve this by making the substitution u=\sin(x) so du=\cos(x)dx . Then we can write the whole integrand in terms of u by using the identity

\cos^2(x)=1-\sin^2(x)=1-u^2 .

So

\int\cos^3(x)\sin^2(x)dx =\int\cos^2(x)\sin^2(x)\cos(x)dx
=\int (1-u^2)u^2du
=\int u^2du-\int u^4du
=\frac{u^3}{3}+\frac{u^5}{5}+C
=\frac{\sin^3(x)}{3}-\frac{\sin^5(x)}{5}+C

This method works whenever there is an odd power of sine or cosine.

To evaluate \int\cos^m(x)\sin^n(x)dx when either m or n is odd.

  • If m is odd substitute u=\sin(x) and use the identity \cos^2(x)=1-\sin^2(x)=1-u^2 .
  • If n is odd substitute u=\cos(x) and use the identity \sin^2(x)=1-\cos^2(x)=1-u^2 .

ExampleEdit

Find \int\limits_0^\frac{\pi}{2} \cos^{40}(x)\sin^3(x)dx .

As there is an odd power of \sin we let u=\cos(x) so du=-\sin(x)dx . Notice that when x=0 we have u=\cos(0)=1 and when x=\frac{\pi}{2} we have u=\cos(\tfrac{\pi}{2})=0 .

\int\limits_0^\frac{\pi}{2} \cos^{40}(x)\sin^3(x)dx =\int\limits_0^\frac{\pi}{2} \cos^{40}(x)\sin^2(x)\sin(x)dx
=-\int\limits_1^0 u^{40}(1-u^2)du
=\int\limits_0^1 u^{40}(1-u^2)du
=\int\limits_0^1(u^{40}-u^{42})du
=\left(\frac{u^{41}}{41}-\frac{u^{43}}{43}\right)\Bigg|_0^1
=\frac{1}{41}-\frac{1}{43}

When both m and n are even things get a little more complicated.

To evaluate \int\cos^m(x)\sin^n(x)dx when both m and n are even.


Use the identities \sin^2(x)=\frac{1-\cos(2x)}{2} and \cos^2(x)=\frac{1+\cos(2x)}{2} .

ExampleEdit

Find \int\sin^2(x)\cos^4(x)dx .

As \sin^2(x)=\frac{1-\cos(2x)}{2} and \cos^2(x)=\frac{1+\cos(2x)}{2} we have

\int\sin^2(x)\cos^4(x)dx=\int\left(\frac{1-\cos(2x)}{2}\right)\left(\frac{1+\cos(2x)}{2}\right)^2dx

and expanding, the integrand becomes

\frac{1}{8}\int\left(1-\cos^2(2x)+\cos(2x)-\cos^3(2x)\right)dx

Using the multiple angle identities

I =\frac{1}{8}\left(\int 1dx-\int\cos^2(2x)dx+\int\cos(2x)dx-\int\cos^3(2x)dx\right)
=\frac{1}{8}\left(x-\frac{1}{2}\int\Big(1+\cos(4x)\Big)dx+\frac{\sin(2x)}{2}-\int\cos^2(2x)\cos(2x)dx\right)
=\frac{1}{16}\left(x+\sin(2x)+\int\cos(4x)dx-2\int\Big(1-\sin^2(2x)\Big)\cos(2x)dx\right)

then we obtain on evaluating

I=\frac{x}{16}-\frac{\sin(4x)}{64}+\frac{\sin^3(2x)}{48}+C

Powers of Tan and SecantEdit

To evaluate \int\tan^m(x)\sec^n(x)dx .

  1. If n is even and n\ge 2 then substitute u=tan(x) and use the identity \sec^2(x)=1+\tan^2(x) .
  2. If n and m are both odd then substitute u=\sec(x) and use the identity \tan^2(x)=\sec^2(x)-1 .
  3. If n is odd and m is even then use the identity \tan^2(x)=\sec^2(x)-1 and apply a reduction formula to integrate \sec^j(x)dx , using the examples below to integrate when j=1,2 .

Example 1Edit

Find \int\sec^2(x)dx .

There is an even power of \sec(x) . Substituting u=\tan(x) gives du=\sec^2(x)dx so

\int\sec^2(x)dx=\int du=u+C=\tan(x)+C.


Example 2Edit

Find \int\tan(x)dx .

Let u=\cos(x) so du=-\sin(x)dx . Then

\int\tan(x)dx =\int\frac{\sin(x)}{\cos(x)}dx
=\int -\frac{du}{u}
=-\ln|u|+C
=-\ln\Big|\cos(x)\Big|+C
=\ln\Big|\sec(x)\Big|+C


Example 3Edit

Find \int\sec(x)dx .

The trick to do this is to multiply and divide by the same thing like this:

\int\sec(x)dx =\int\sec(x)\frac{\sec(x)+\tan(x)}{\sec(x)+\tan(x)}dx
=\int\frac{\sec^2(x)+\sec(x)\tan(x)}{\sec(x)+\tan(x)}

Making the substitution u=\sec(x)+\tan(x) so du=\sec(x)\tan(x)+\sec^2(x)dx ,

\int\sec(x)dx =\int\frac{du}{u}
=\ln|u|+C
\ln\Big|\sec(x)+\tan(x)\Big|+C

More trigonometric combinationsEdit

For the integrals \int\sin(nx)\cos(mx)dx or \int\sin(nx)\sin(mx)dx or \int\cos(nx)\cos(mx)dx use the identities

  • \sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2}
  • \sin(a)\sin(b)=\frac{\cos(a-b)-\cos(a+b)}{2}
  • \cos(a)\cos(b)=\frac{\cos(a-b)+\cos(a+b)}{2}

Example 1Edit

Find \int\sin(3x)\cos(5x)dx .

We can use the fact that \sin(a)\cos(b)=\frac{\sin(a+b)+\sin(a-b)}{2} , so

\sin(3x)\cos(5x)=\frac{\sin(8x)+\sin(-2x)}{2}

Now use the oddness property of \sin(x) to simplify

\sin(3x)\cos(5x)=\frac{\sin(8x)-\sin(2x)}{2}

And now we can integrate

\int\sin(3x)\cos(5x)dx =\int\Big(\sin(8x)-\sin(2x)\Big)dx
=\frac{\cos(2x)}{4}-\frac{\cos(8x)}{16}+C


Example 2Edit

Find:\int\sin(x)\sin(2x)dx .

Using the identities

\sin(x)\sin(2x)=\frac{\cos(-x)-\cos(3x)}{2}=\frac{\cos(x)-\cos(3x)}{2}

Then

\int\sin(x)\sin(2x)dx =\frac{1}{2}\int\Big(\cos(x)-\cos(3x)\Big)dx
=\frac{\sin(x)}{2}-\frac{\sin(3x)}{6}+C
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Integration techniques/Trigonometric Integrals