# Evaluating Trigonometric Integral

1. Jun 5, 2014

### jdawg

1. The problem statement, all variables and given/known data
$\int$$\frac{\pi}{2}$0 sin7y dy
The bounds are from $\frac{\pi}{2}$ to 0.

2. Relevant equations

3. The attempt at a solution
I think I did the integration correctly, but I don't really know how to evaluate this.

$\int$siny(sin2y)3 dy
$\int$siny(1-cos2y)3 dy
u=cosy
du=-siny dy

-$\int$(1-u2)3 du
-$\int$(-u6+3u4-3u3+1) du

$\frac{u7}{7}$-$\frac{3u5}{5}$+$\frac{3u4}{4}$-u

$\frac{cos7y}{7}$-$\frac{3cos5y}{5}$+$\frac{3cos4y}{4}$-cosy

2. Jun 5, 2014

### Ray Vickson

No the bounds are not from $\pi/2$ to 0, they are from 0 to $\pi/2$. Anyway, you have the right idea; you just need to figure out the bounds after the u-substitution; that is, what are $a$ and $b$ in
$$\int_0^{\pi/2} \sin^7 y \: dy = -\int_a^b (1-u^2)^3 \, du?$$

3. Jun 5, 2014

### jdawg

Oops! That's what I meant. Haha thanks, I guess it would be a lot easier to just change the bounds.

4. Jun 5, 2014

### jdawg

Hey! I ran into another problem, I changed the bounds to 1 to 0:

-∫ (-u6+3u4-3u3+1) du

Then I flipped the bounds to 0 to 1 and changed the sign on the integral:

∫(-u6+3u4-3u3+1) du

Then I plugged in my bounds and got 99/140, but the answer in the back of the book is 16/35. I'm having trouble figuring out what I did wrong.

5. Jun 5, 2014

### Pranav-Arora

How do you get $u^3$? It should be $u^2$ instead.

6. Jun 6, 2014

### jdawg

Ooops! It was just a silly mistake. Thanks so much!