Trig Integral sin^2 x + cos^2 x = 1

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Homework Statement



[itex]\int[/itex]sin[itex]^{5}[/itex]x cos x dx

Homework Equations



sin[itex]^{2}[/itex]x + cos[itex]^{2}[/itex] x = 1

The Attempt at a Solution



I've at least written down that sin[itex]^{5}[/itex]x = (sin[itex]^{2}[/itex]x)[itex]^{2}[/itex] sin x. Then I set sin[itex]^{2}[/itex]x equal to 1 - cos[itex]^{2}[/itex]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 [itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

The answer I received was [itex]\frac{1}{6}[/itex]sin[itex]^{6}[/itex]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[itex]^{2}[/itex]x)
 
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Youngster said:

Homework Statement



[itex]\int[/itex]sin[itex]^{5}[/itex]x cos x dx

Homework Equations



sin[itex]^{2}[/itex]x + cos[itex]^{2}[/itex] x = 1

The Attempt at a Solution



I've at least written down that sin[itex]^{5}[/itex]x = (sin[itex]^{2}[/itex]x)[itex]^{2}[/itex] sin x. Then I set sin[itex]^{2}[/itex]x equal to 1 - cos[itex]^{2}[/itex]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 [itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

The answer I received was [itex]\frac{1}{6}[/itex]sin[itex]^{6}[/itex]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[itex]^{2}[/itex]x)
Do a u substitution directly on the problem as it was initially given to you.

What is the derivative of sin(x) ?
 
...

Right, I feel kinda "slow" now.

Any way to use the trig identity to solve this though?
 
Zondrina said:
... Make the substitution u = sinx and find du just like sammy said. The rest will follow.

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

Yes, you can do it that way as well. Though I think there's a sign error in the answer you gave.
 
-[itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

I think I was missing that negative. Is this correct?
 
Youngster said:
-[itex]\frac{1}{6}[/itex]cos[itex]^{6}[/itex]x + [itex]\frac{1}{2}[/itex]cos[itex]^{4}[/itex]x - [itex]\frac{1}{2}[/itex]cos[itex]^{2}[/itex]x + C

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

Yep.