Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int x \sin x \cos 2 x d x$

Solution:

Let $I=\int x \sin x \cos 2 x d x=\frac{1}{2} \int x \times 2 \sin x \cos 2 x d x$

Using integration by parts,

$=\frac{1}{2} \int x(\sin (x+2 x)-\sin (2 x-x)) d x$

$=\frac{1}{2} \int x(\sin 3 x-\sin x) d x$

Using integration by parts,

$=\frac{1}{2}\left(x \int(\sin 3 x-\sin x) d x-\int \frac{d}{d x} x \int(\sin 3 x-\sin x) d x\right) d x$

$=\frac{1}{2}\left[x\left(\frac{-\cos 3 x}{3}+\cos x\right)-\int-\left(\frac{\cos 3 x}{3}+\cos x\right) d x\right)$

$I=\frac{1}{2}\left[-x \frac{\cos 3 x}{3}+x \cos x+\frac{1}{9} \sin 3 x-\sin x\right]+c$

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