Sin[x]sin[2x]x dx again.

Main Question or Discussion Point

$$\int^a_b x\sin{x}\sin{2x} dx$$

answers? I tried to solve once, it took like 3 pages.

Hurkyl
Staff Emeritus
Gold Member
To solve the integral $$\mbox{\int x\sin x \sin 2x \,dx}$$, let's first solve $$\mbox{\int \sin x \sin 2x \,dx.}$$

$$\begin{equation*} \begin{split} \int \sin{x}\sin{2x} &= \int 2 \sin^2 x \cos x\,dx \\ &= \int 2 u^2 du \quad (u = \sin x) \\ &= \frac{2}{3} u^3 + C\\ &= \frac{2}{3} \sin^3 x + C \end{split} \end{equation*}$$

Now, to solve $$\mbox{\int x\sin x \sin 2x \,dx}$$ we may now do integration by parts via

$$\begin{equation*} \begin{split} u &= x \\ dv &= \sin x \sin 2x \,dx \\ du &= dx \\ v &= \frac{2}{3} \sin^3 x \end{split} \end{equation*}$$

Which yields:

$$\begin{equation*} \begin{split} \int x\sin x \sin 2x \,dx &= x \frac{2}{3} \sin^3 x - \int \frac{2}{3} \sin^3 x \,dx \\ &= \frac{2}{3} x \sin^3 x - \frac{2}{3} \int \sin x (1 - \cos^2 x)\,dx \\ &= \frac{2}{3} x \sin^3 x - \frac{2}{3} \left( \int \sin x \, dx - \int \sin x \cos^2 x \, dx \right) \\ &= \frac{2}{3} x \sin^3 x - \frac{2}{3}\left( -\cos x + \frac{1}{3} \cos^3 x \right) + C \\ &= \frac{2}{3} x \sin^3 x + \frac{2}{3} \cos x - \frac{2}{9} \cos^3 x + C \end{split} \end{equation*}$$

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Amazing. I have learned yet a new way from you.

Out of curiousity, what background art thou?

Hurkyl
Staff Emeritus