- #1

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## 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?