Solving ∫x sin^3x dx with Integration by Parts

[Bimpo]

Homework Statement

Finding ∫x sin^3x dx

Homework Equations

I don't think this is needed

The Attempt at a Solution

∫x sin^3x dx using integration by parts u = x, du = dx, dv = sin^3x, v = 1/3cos^3x - cosx
= (x)(1/3cos^3x - cosx) - ∫1/3cos^3x - ∫cosx dx
= (x)(1/3cos^3x - cosx) - 1/3∫cos^3x dx + (sinx)
= (x)(1/3cos^3x - cosx) - 1/3∫cos^2xcosx dx + (sinx)
= (x)(1/3cos^3x - cosx) - 1/3∫(1-sin^2x)(cosx) dx + (sinx)
= (x)(1/3cos^3x - cosx) - (1/3∫cosx dx - ∫cosxsin^2x dx) + (sinx)
= (x)(1/3cos^3x - cosx) - (1/3 (-sinx) - ∫cosxsin^2x dx) + (sinx)
using subsitution u = sinx, du = cosx dx
= (x)(1/3cos^3x - cosx) - (1/3 (-sinx) - ∫u^2 du) + (sinx)
= (x)(1/3cos^3x - cosx) - (1/3 (-sinx) - 1/3u^3) + (sinx)
then subsitute back
= (x)(1/3cos^3x - cosx) - (1/3 (-sinx) - 1/3sin^3x) + (sinx) + C

Anything I did wrong?
Anyway to do this much easier/faster?

 

[SithsNGiggles]

Don't integrate by parts. Not yet, at least.

Use the half-angle identity:
sin²x = 1/2(1 - cos(2x))

 

[Bimpo]

Ok, I do that then
I'll get something like:

= 1/2∫(1-cos2x)xsinx dx
= 1/2∫xsinx - ∫xsinxcos2x dx <-- I get stuck here

 

[vela]

You can always check your answer by differentiating it and seeing if you recover the integrand.

 

[SammyS]

Homework Statement

Finding ∫x sin^3x dx

The Attempt at a Solution

∫x sin^3x dx using integration by parts u = x, du = dx, dv = sin^3x  dx, v = 1/3cos^3x - cosx

To find v by integrating, I would rewrite sin³ x as

sin³ x = (1 - cos² x)sin  x

= sin  x - sin  x   cos² x​

This is fairly easy to integrate.