# Finding ∫x sin^3x dx

## 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?

Related Calculus and Beyond Homework Help News on Phys.org
Don't integrate by parts. Not yet, at least.

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

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

SammyS
Staff Emeritus
Homework Helper
Gold Member

## 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.