Question

solve this question..and don't write answer directly​

Answer 1

We need to evaluate,

$\displaystyle\longrightarrow I=\int x\,\sin x\,\sin(2x)\,\sin(3x)\ dx$

or,

$\displaystyle\longrightarrow I=\int x\,\sin(2x)\left[\sin x\,\sin(3x)\right]\ dx$

Recall,

• $\sin A\sin B=\dfrac{1}{2}\left[\cos(A-B)-\cos(A+B)\right]$

Taking $A=x$ and $B=3x,$

• $\sin x\sin(3x)=\dfrac{1}{2}\left[\cos(x-3x)-\cos(x+3x)\right]$

Then,

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int x\,\sin(2x)\left[\cos(x-3x)-\cos(x+3x)\right]\ dx$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int x\,\sin(2x)\left[\cos(2x)-\cos(4x)\right]\ dx\quad\quad\left[\,\cos(-x)=\cos x\,\right]$

$\displaystyle\longrightarrow I=\dfrac{1}{2}\int x\,\sin(2x)\cos(2x)\ dx-\dfrac{1}{2}\int x\,\sin(2x)\cos(4x)\ dx\quad\quad\dots(1)$

Let,

$\displaystyle\longrightarrow I_1=\int x\,\sin(2x)\,\cos(2x)\ dx$

Recall,

• $\sin A\cos A=\dfrac{1}{2}\,\sin(2A)$

Then,

$\displaystyle\longrightarrow I_1=\dfrac{1}{2}\int x\,\sin(4x)\ dx$

Integrating by parts,

$\displaystyle\longrightarrow I_1=\dfrac{1}{2}\left[-\dfrac{x}{4}\cos(4x)+\dfrac{1}{4}\int\cos(4x)\ dx\right]$

$\displaystyle\longrightarrow I_1=\dfrac{1}{32}\sin(4x)-\dfrac{x}{8}\cos(4x)$

Let,

$\displaystyle\longrightarrow I_2=\int x\,\sin(2x)\,\cos(4x)\ dx$

Recall,

• $\sin A\cos B=\dfrac{1}{2}\left[\sin(A+B)+\sin(A-B)\right]$

Then,

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\int x\left[\sin(2x+4x)+\sin(2x-4x)\right]\ dx$

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\int x\left[\sin(6x)-\sin(2x)\right]\ dx\quad\quad\left[\,\sin(-x)=-\sin x\,\right]$

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\int x\,\sin(6x)\ dx-\dfrac{1}{2}\int x\,\sin(2x)\ dx\quad\quad\dots(2)$

Let,

$\displaystyle\longrightarrow I_{21}=\int x\,\sin(6x)\ dx$

Integrating by parts,

$\displaystyle\longrightarrow I_{21}=-\dfrac{x}{6}\,\cos(6x)+\dfrac{1}{6}\int\cos(6x)\ dx$

$\displaystyle\longrightarrow I_{21}=\dfrac{1}{36}\sin(6x)-\dfrac{x}{6}\,\cos(6x)$

Let,

$\displaystyle\longrightarrow I_{22}=\int x\,\sin(2x)\ dx$

Integrating by parts,

$\displaystyle\longrightarrow I_{22}=-\dfrac{x}{2}\,\cos(2x)+\dfrac{1}{2}\int\cos(2x)\ dx$

$\displaystyle\longrightarrow I_{22}=\dfrac{1}{4}\sin(2x)-\dfrac{x}{2}\,\cos(2x)$

Then (2) becomes,

$\displaystyle\longrightarrow I_2=\dfrac{1}{2}\left(\dfrac{1}{36}\sin(6x)-\dfrac{x}{6}\cos(6x)\right)-\dfrac{1}{2}\left(\dfrac{1}{4}\sin(2x)-\dfrac{x}{2}\cos(2x)\right)$

$\displaystyle\longrightarrow I_2=\dfrac{1}{72}\sin(6x)-\dfrac{x}{12}\cos(6x)-\dfrac{1}{8}\sin(2x)+\dfrac{x}{4}\cos(2x)\right)$

Hence (1) becomes,

$\displaystyle\begin{aligned}\longrightarrow I=\ \ &\dfrac{1}{2}\left(\dfrac{1}{32}\sin(4x)-\dfrac{x}{8}\cos(4x)\right)\\-\ \ &\dfrac{1}{2}\left(\dfrac{1}{72}\sin(6x)-\dfrac{x}{12}\cos(6x)-\dfrac{1}{8}\sin(2x)+\dfrac{x}{4}\cos(2x)\right)+C\end{aligned}$

$\displaystyle\small\text{ \longrightarrow\underline{\underline{I=\dfrac{1}{8}\left(\dfrac{1}{8}\sin(4x)-\dfrac{x}{2}\cos(4x)-\dfrac{1}{18}\sin(6x)+\dfrac{x}{3}\cos(6x)+\dfrac{1}{2}\sin(2x)-x\cos(2x)\right)+C}} }$