Integration by parts I believe

[afcwestwarrior]

Homework Statement

∫x sin^2 x dx

Homework Equations

integration by parts ∫u dv= uv-∫ v du

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]

[tex]cos2x = 2cos^2x-1 = 1-2sin^2x=cos^2x-sin^2x[/tex]

I believe you mean [tex]sin^2x = 1 - cos^2 x[/tex]?

 

[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 [tex]\sin^2 x = \frac{1}{2}(1-\cos (2x))[/tex].