Integration by parts I believe

[afcwestwarrior]

1. The problem statement, all variables and given/known data
∫x sin^2 x dx



2. Relevant equations

integration by parts ∫u dv= uv-∫ v du

3. The attempt at a solution
u=x dv=1-cos2x
v= 1/2 sin 2x
du=dx

is that correct

i substituted sin^2 x= 1-cos2x Am I allowed to do that.

[konthelion]

cos2x = 2cos^2x-1 = 1-2sin^2x=cos^2x-sin^2x

I believe you mean sin^2x = 1 - cos^2 x?

[afcwestwarrior]

oh ok I understand.

[afcwestwarrior]

So it would be ∫x (1-cos^2x) dx

and then i'd subsitute u= cos x
du=-sin x

so then it would be ∫x- u^2 x^2 dx

is that correct

[Defennder]

You should use this identity \sin^2 x = \frac{1}{2}(1-\cos (2x)).