# Thread: Integration of trigonometric functions

1. ## Integration of trigonometric functions

Hi,

I'm currently studying integration in preperation for an end of module test. So far it's been going well however, I'm finding the section on Integration of trigonometric functions a bit perplexing! The examples in my notes I can follow quite easily and I understand them, and I have got some books from the library and what's written in them I can understand when I follow them too. When it comes to doing questions by myself though, I'm finding it a bit tough!

The first question is:

Integrate the following:

[integral] sin^4x cosx dx,

I know that you have to use the identity - sin^2 x = 1/2(1-cos(2x)). That's easy to do if it's to the power of 2 however as this is a power of 4, I'm not quite sure what to do with it :/

I did try the question and the answer I got was just way way way off the mark, the actual answer is 1/2 sin^5 x + C ........

I wonder if anyone could guide me as to what I am supposed to do with it

cheers!

2. Originally Posted by scott.b89
Hi,

I'm currently studying integration in preperation for an end of module test. So far it's been going well however, I'm finding the section on Integration of trigonometric functions a bit perplexing! The examples in my notes I can follow quite easily and I understand them, and I have got some books from the library and what's written in them I can understand when I follow them too. When it comes to doing questions by myself though, I'm finding it a bit tough!

The first question is:

Integrate the following:

[integral] sin^4x cosx dx,

I know that you have to use the identity - sin^2 x = 1/2(1-cos(2x)). That's easy to do if it's to the power of 2 however as this is a power of 4, I'm not quite sure what to do with it :/

I did try the question and the answer I got was just way way way off the mark, the actual answer is 1/2 sin^5 x + C ........

I wonder if anyone could guide me as to what I am supposed to do with it

cheers!
When dealing with integrals of the form

$\displaystyle \int \sin^m x \cos^n x dx$

when m and n are integers, 4 cases arise:

(1) m and n are odd
(2) m is even and n odd
(3) m is odd and n even
(4) m and n are even

(2) m is even and n odd
Pull off 1 $\displaystyle \cos x$ term and make the substitution $\displaystyle u = \sin x$

(3) m is odd and n even
Pull off 1 $\displaystyle \sin x$ term and make the substitution $\displaystyle u = \cos x$

(1) m and n are odd
This is case (2) and (3) so do either.

(4) m and n are even
Use the double angle formulas

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

Let's look at a version of your problem
$\displaystyle \int \sin^4x \cos^3 x dx$
Here the $\displaystyle \cos x$ term is odd so pull off a $\displaystyle \cos x$ term
$\displaystyle \int \sin^4x \cos^3 x dx = \int \sin^4x \cos^2 x \cos xdx$
and let $\displaystyle u = \sin x$ so $\displaystyle du = \cos x\,dx$ and the integral becomes
$\displaystyle \int u^4(1-u^2)du$
noting that I used the identity $\displaystyle \cos^2x = 1 - \sin^2x.$

3. Big help!

Thanks very much mate