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

[Youngster]

Homework Statement

\intsin^{5}x cos x dx

Homework Equations

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

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)

 

[SammyS]

Homework Statement

\intsin^{5}x cos x dx

Homework Equations

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

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)

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

What is the derivative of sin(x) ?

 

[Youngster]

...

Right, I feel kinda "slow" now.

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

 

[STEMucator]

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

 

[Youngster]

... 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

 

[Dick]

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.

 

[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?

 

[Dick]

-\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?

Yep.