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.

Powers of Sine and Cosine[edit]

We will give a general method to solve generally integrands of the form \cos^m(x)\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

\begin{matrix}
\int\cos^3(x)\sin^2(x)\,dx &=&\int\cos^2(x)\sin^2(x)\cos(x)\,dx\\
&=&\int (1-u^2)u^2\,du\\
&=&\int u^2\,du - \int u^4\,du\\
&=&{1\over 3} u^3+{1\over 5}u^5 + C\\
&=&{1\over 3} \sin^3(x)-{1\over 5}\sin^5(x)+C 
\end{matrix}.

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.

Example[edit]

Find \int_0^{\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=\pi/2 we have u=\cos(\pi/2) = 0.

\begin{matrix}
\int_0^{\pi/2} \cos^{40}(x)\sin^3(x) dx &=&  \int_0^{\pi/2} \cos^{40}(x)\sin^2(x) \sin(x) dx \\ 
&=& -\int_{1}^{0} u^{40} (1-u^2) du \\
&=&\int_{0}^{1} u^{40} (1-u^2) du\\
&=& \int_{0}^{1} u^{40} - u^{42} du \\
&=& [\frac{1}{41}u^{41} - \frac{1}{43}u^{43}]_0^1 \\
&=& \frac{1}{41}-\frac{1}{43}.
\end{matrix}

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

Example[edit]

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

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


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

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

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

then we obtain on evaluating

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

Powers of Tan and Secant[edit]

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 1[edit]

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 2[edit]

Find \int \tan(x)dx.

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

\begin{matrix}
\int \tan(x)dx &=& \int \frac{\sin(x)}{\cos(x)} dx \\
&=& \int \frac{-1}{u} du \\
&=& -\ln |u| + C \\
&=& -\ln |\cos(x) | + C\\
&=& \ln |\sec(x)| +C.
\end{matrix}


Example 3[edit]

Find \int \sec(x)dx.

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

\begin{matrix}
\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)}
\end{matrix}.

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

\begin{matrix}
\int \sec(x) dx &=& \int \frac{1}{u} du\\
&=& \ln |u| + C \\
&=& \ln |\sec(x) + \tan(x)| + C
\end{matrix}.

More trigonometric combinations[edit]

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) = {1\over 2}(\sin{(a+b)}+\sin{(a-b)}) \,
  •  \sin(a)\sin(b) = {1\over 2}(\cos{(a-b)}-\cos{(a+b)}) \,
  •  \cos(a)\cos(b) = {1\over 2}(\cos{(a-b)}+\cos{(a+b)}) \,

Example 1[edit]

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

We can use the fact that \sin(a)\cos(b)=(1/2)(\sin(a+b)+\sin(a-b)), so

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

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

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

And now we can integrate

\begin{matrix}
\int \sin(3x)\cos(5x)\,dx & = & \frac{1}{2} \int \sin(8x)-\sin(2x)dx \\
& = &  \frac{1}{2}(-\frac{1}{8}\cos(8x)+\frac{1}{2}\cos(2x)) +C \\
\end{matrix}

Example 2[edit]

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

Using the identities

\sin(x) \sin(2x)= \frac{1}{2} \left( \cos(-x)-\cos(3x) \right)
= \frac{1}{2} (\cos(x) -\cos(3x)).

Then

\begin{matrix} 
\int \sin(x)\sin(2x)\,dx & = & \frac{1}{2} \int (\cos(x)-\cos(3x))\,dx  \\
& = &  \frac{1}{2}(\sin(x)-\frac{1}{3}\sin(3x)) + C
\end{matrix}
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Integration techniques/Trigonometric Integrals