What is the best way?
Idk, you didn't show any work.
I guess I'll have to use an identity. Maybe sin^2(x) = 1-cos^2(x)?
u = 1- cos^2(x)
du/dx = -2cos(x)sin(x)
(1-cos^2(x))*sin(x) dx = (-u sin(x)/2cos(x)) du
Doesn't really help, or?
[tex]\int\cos x\cos^2 xdx[/tex]
Use a BASIC trig identity to change the 2nd degree cosine function.
not cosine, sine
What are you talking about?
Don't know, I guess I'm too drunk to do maths right now.
Try again later :)
Oh, I wrote cos^3 x instead of sin^3 x in the headline. That explains my confusion.
sin^2 x*sin x
(1 - cos^2 x)sin x
u = 1-cos^2 x
du/dx = 2cos x*sin*x
sin^3 x dx = - u / 2cos x
distribute the sinx and you'll see your solution
So how about just u= cos(x)?
You know what they say "Don't drink and derive"!
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