Question

Evaluate $\int (x\cdot \cos x\cdot \cos 2x)dx$
(where $C$ is constant of integration)

• A. $x(\sin x-\frac{2}{3}{\sin }^{3}x)+\frac{\cos x}{2}-\frac{\cos 3x}{18}+C$
• B. $x(\sin x+\frac{2}{3}{\sin }^{3}x)+\frac{\cos x}{2}-\frac{\cos 3x}{18}+C$
• C. $x(\sin x-\frac{2}{3}{\sin }^{3}x)+\frac{\cos x}{2}+\frac{\cos 3x}{18}+C$
• D. $x(\sin x+\frac{2}{3}{\sin }^{3}x)-\frac{\cos x}{2}+\frac{\cos 3x}{18}+C$

Answer 1

The correct option is C $x(\sin x-\frac{2}{3}{\sin }^{3}x)+\frac{\cos x}{2}+\frac{\cos 3x}{18}+C$

Let $I=\int (x\cdot \cos x\cdot \cos 2x)dx$
$\Rightarrow I=\int x\cos x(1-2{\sin }^{2}x)dx\cdots \left(i\right)$

Let $I_1=\int \cos x(1-2{\sin }^{2}x)dx$
Put $\sin x=t\Rightarrow \cos xdx=dt$
$\Rightarrow I_1=\int (1-2t^2)dt$
$\therefore I_1=t-\frac{2t^3}{3}+C$
$\Rightarrow I_1=\sin x-\frac{2{\sin }^{3}x}{3}+C_1$

Now using ILATE in $\left(i\right),$ we get
$I=x(\sin x-\frac{2}{3}{\sin }^{3}x)-\int (\sin x-\frac{2}{3}{\sin }^{3}x)dx$
$\Rightarrow I=x(\sin x-\frac{2}{3}{\sin }^{3}x)+\cos x+\frac{2}{3}\int {\sin }^{3}xdx$

Let $I_2=\int {\sin }^{3}xdx=\int \left(\frac{3\sin x-\sin 3x}{4}\right)dx$
$(\because \sin 3x=3\sin x-4{\sin }^{3}x)$
$\therefore I_2=-\frac{3}{4}\cos x+\frac{\cos 3x}{12}+C_3$

$\Rightarrow I=x(\sin x-\frac{2}{3}{\sin }^{3}x)+\cos x-\frac{\cos x}{2}+\frac{\cos 3x}{18}+C$

$\therefore I=x(\sin x-\frac{2}{3}{\sin }^{3}x)+\frac{\cos x}{2}+\frac{\cos 3x}{18}+C$