# Integrals - the Substitution Rule with sin^n(x)

1. Sep 11, 2012

### sugarxsweet

1. The problem statement, all variables and given/known data
Given that n is a positive integer, prove ∫sin^n(x)dx=∫cos^n(x)dx from 0 -> pi/2

2. Relevant equations
Perhaps sin^2(x)+cos^2(x)=1? Not sure.

3. The attempt at a solution
I honestly don't even know where to start. Should I set u=sin(x) or cos(x)? Doesn't seem to get the right answer either way...

2. Sep 11, 2012

### Simon Bridge

You need to show that the area under the graph from 0 to 90deg is the same for sin and cosine to any power.

Your first step is to understand the problem - try sketching the graphs for a few powers and shading the area in question to see what you are up against.

There is a rule for integrating powers of sin and cos ... you can derive it from integration by parts. Knowing the rule, you can probably just do it algebraically.

Or you can try a substitution in one like $x=\frac{\pi}{2}-u$ and exploit the parity of the functions.

3. Sep 11, 2012

### sugarxsweet

Sorry, to clarify, the hint says:
Use a trigonometric identity and substitution. Do not solve the deﬁnite integrals

Given this information, how would you recommend solving?

4. Sep 12, 2012

### SammyS

Staff Emeritus
The co-function identity states that $\displaystyle \cos(x) = \sin\left(\frac{\pi}{2}-x\right)\ .$

5. Sep 12, 2012

### sugarxsweet

Thanks! I got so far:

∫cos^n(x)dx from 0 to pi/2
=∫sin^n(pi/2-x)dx from 0 to pi/2
=∫sin^n(-x)dx from -pi/2 to 0
=∫sin^n(x)dx from 0 to pi/2

Does this look right to you? Thanks for the hint!!

6. Sep 12, 2012

### Simon Bridge

Yeah, that's the idea - it was the second hint post #2 :)

You should comment each step in your actual answer, to explain what you are doing.

7. Sep 12, 2012

### SammyS

Staff Emeritus
Use the substitution, $u=\frac{\pi}{2}-x$ to show that $\displaystyle \int_{x=0}^{x=\pi/2}\sin^n(\frac{\pi}{2}-x)\,dx = \int_{u=\pi/2}^{u=0}-\sin(u)\,du\ .$

8. Sep 12, 2012

Thanks!