Integrate sinx(cosx)^2dx using the substitution u=sinx. I know how to do this using u =cosx, but not sinx. The next problem on the homework was the same question except it asked to use u=cosx, so there couldn't have been a mistake.

Answer 1

u = sinx
du = cosx dx
so, dx = du/cosx = du/√(1-u^2)

sinx(cosx)^2 dx = u(1-u^2)/√(1-u^2) du
= u/√(1-u^2) du

Now, let v = √(1-u^2)
dv = -u/√(1-u^2) du

and you have

-v dv

integrate that to get

-1/2 v^2 = -1/2(1-u^2) = -1/2 (1-sin^2(x)) = -1/2 cos^2(x) + C

Now, that probably looks different from what you got letting u=cosx, but I think if you manipulate things, you'll find that the difference is caused by having a different +C at the end.

Or, maybe I made a mistake above...

Answer 2

ok, I got it, it worked out for me to be the same thing as when I used u=cosx