# Trig Integral

1. Oct 17, 2012

### Youngster

1. The problem statement, all variables and given/known data

$\int$sin$^{5}$x cos x dx

2. Relevant equations

sin$^{2}$x + cos$^{2}$ x = 1

3. The attempt at a solution

I've at least written down that sin$^{5}$x = (sin$^{2}$x)$^{2}$ sin x. Then I set sin$^{2}$x equal to 1 - cos$^{2}$x.

I then did a u-substitution, setting u equal to cos x to remove the sin x, and preceeded to integrate with respect to u. I ended up with $\frac{1}{6}$cos$^{6}$x + $\frac{1}{2}$cos$^{4}$x - $\frac{1}{2}$cos$^{2}$x + C

The answer I received was $\frac{1}{6}$sin$^{6}$x + C

Can I get some insight on how to obtain that?

My textbook does give a procedure for this specific case (where one of the powers is odd), but I don't get the following instructions: "Then we combine the single sin x with dx in the integral and set sinxdx equal to -d(cosx)" which is stated after using the identity to replace the (sin$^{2}$x)

2. Oct 17, 2012

### SammyS

Staff Emeritus
Do a u substitution directly on the problem as it was initially given to you.

What is the derivative of sin(x) ?

3. Oct 17, 2012

### Youngster

...

Right, I feel kinda "slow" now.

Any way to use the trig identity to solve this though?

4. Oct 17, 2012

### Zondrina

.... Make the substitution u = sinx and find du just like sammy said. The rest will follow.

5. Oct 17, 2012

### Youngster

Yeah, I did that. Just curious if I can still use the identity to solve this though. Otherwise, I suppose that's it

6. Oct 17, 2012

### Dick

Yes, you can do it that way as well. Though I think there's a sign error in the answer you gave.

7. Oct 17, 2012

### Youngster

-$\frac{1}{6}$cos$^{6}$x + $\frac{1}{2}$cos$^{4}$x - $\frac{1}{2}$cos$^{2}$x + C

I think I was missing that negative. Is this correct?

8. Oct 17, 2012

Yep.