Integrate ∫xcos⁡(x)sin⁡(x)dxI've been trying to integrate by parts, but I can't! I know there's a trigonometric function about cos⁡(2x) but I don't know how to integrate that function wit sin⁡x

Given∫xcos⁡(x)sin⁡(x) dxUse the identity cos⁡(x)sin⁡(x)=sin⁡(2x)2Now∫x sin⁡(2x)2 dx=12∫xsin⁡(2x) dxNow use Integration By Parts u=x and v′=sin⁡(2x)

∫xcos⁡(x)sin⁡(x)dx=∫x12sin⁡(2x)dx=12∫xsin⁡(2x)dx=12x[−12cos⁡(2x)]−12∫[−12cos⁡(2x)]dx+C=−14xcos⁡(2x)+18∫cos⁡(2x)d(2x)+C=−14xcos⁡(2x)+18sin⁡(2x)+C

Given∫xcos⁡(x)sin⁡(x) dxUse the identity cos⁡(x)sin⁡(x)=sin⁡(2x)2Now∫x sin⁡(2x)2 dx=12∫xsin⁡(2x) dxNow use Integration By Parts u=x and v′=sin⁡(2x)

∫xcos⁡(x)sin⁡(x)dx=∫x12sin⁡(2x)dx=12∫xsin⁡(2x)dx=12x[−12cos⁡(2x)]−12∫[−12cos⁡(2x)]dx+C=−14xcos⁡(2x)+18∫cos⁡(2x)d(2x)+C=−14xcos⁡(2x)+18sin⁡(2x)+C

Get the solution to "Integrate ∫xcos⁡(x)sin⁡(x)dxI've been trying to integrate by parts,..."

dreangannaa

2022-04-24

Integrate

\int x \cos (x) \sin (x) d x

I've been trying to integrate by parts, but I can't! I know there's a trigonometric function about

\cos (2 x)

but I don't know how to integrate that function wit

\sin x

drenkttj9

Beginner2022-04-25Added 20 answers

Given

\int x \cos (x) \sin (x) d x

Use the identity

\cos (x) \sin (x) = \frac{\sin (2 x)}{2}

Now

\int x \frac{\sin (2 x)}{2} d x = \frac{12}{\int} x \sin (2 x) d x

Now use Integration By Parts u=x and

v^{'} = \sin (2 x)

vanvanvan4ie

Beginner2022-04-26Added 12 answers

\int x \cos (x) \sin (x) d x = \int x \frac{1}{2} \sin (2 x) d x = \frac{1}{2} \int x \sin (2 x) d x

= \frac{1}{2} x [- \frac{1}{2} \cos (2 x)] - \frac{1}{2} \int [- \frac{1}{2} \cos (2 x)] d x + C

= - \frac{1}{4} x \cos (2 x) + \frac{1}{8} \int \cos (2 x) d (2 x) + C

= - \frac{1}{4} x \cos (2 x) + \frac{1}{8} \sin (2 x) + C

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