Question

The value of the integral ${\int }_{-\pi }^{\pi }(\cos ax-\sin bx{)}^{2}dx$, where $a$ and $b$ are integers, is

• A. $2\pi (1+a+b)$
• B. $0$
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is D $2\pi$

$I={\int }_{-\pi }^{\pi }{(\cos ax-\sin bx)}^{2}dx={\int }_{-\pi }^{\pi }{\sin }^{2}bxdx-2{\int }_{-\pi }^{\pi }\left(\left(\cos ax\right)\left(\sin bx\right)\right)dx+{\int }_{-\pi }^{\pi }{\cos }^{2}axdx$
Let $I=I_1+I_2+I_3$ ...(1)
Such that, $I_1={\int }_{-\pi }^{\pi }{\sin }^{2}bxdx$
Substituting $bx=t\Rightarrow bdx=dt$
We get, $I_1=\frac{1}{b}{\int }_{-\pi b}^{\pi b}{\sin }^{2}tdt=\frac{1}{b}{\int }_{-\pi b}^{\pi b}(\frac{1}{2}-\frac{1}{2}\cos 2t)dt$
$=\frac{1}{b}{[\frac{t}{2}-\frac{\sin 2t}{4}]}_{-\pi b}^{\pi b}=\pi -\frac{\sin \left(2\pi b\right)}{2b}$ ...(2)
$I_2=-2{\int }_{-\pi }^{\pi }\left(\left(\cos ax\right)\left(\sin bx\right)\right)dx$
As $\left(\cos ax\right)\left(\sin bx\right)$ is an odd function,
$I_2=0$ ...(3)
$I_3={\int }_{-\pi }^{\pi }{\cos }^{2}axdx$
Substituting, $ax=t\Rightarrow adx=dt$, we get
$I_3=\frac{1}{a}{\int }_{-\pi a}^{\pi a}{\cos }^{2}tdt=\frac{1}{a}{\int }_{-\pi a}^{\pi a}(\frac{1}{2}\cos 2t+\frac{1}{2})dt$
$=\frac{1}{a}{[\frac{t}{2}+\frac{\sin 2t}{4}]}_{-\pi a}^{\pi a}=\frac{\sin \left(2\pi a\right)}{2a}+\pi$ ...(4)
Substituting (2), (3) and (4) in (1), we get
$I=\frac{\sin \left(2\pi a\right)}{2a}-\frac{\sin \left(2\pi b\right)}{2b}+2\pi =0+0+2\pi =2\pi$
Hence, option 'D' is the correct answer.