- #1
Youngster
- 38
- 0
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)