Question

${\int }_{-\pi }^{\pi }(\cos px-\sin qx{)}^{2}dx$ ,where $p,q$ are integers is equal to

• A. $-\pi$
• B. 0
• C. $\pi$
• D. $2\pi$

Answer 1

The correct option is D $2\pi$

Given, ${\int }_{-\pi }^{\pi }[\cos px-\sin qx{]}^{2}dx=I$ (say)
$\Rightarrow$ ${\int }_{-\pi }^{\pi }\left[\cos p\right(-x)-\sin q(-x\left){]}^2\right]dx$ , since $[{\int }_a^{b}F\left(x\right)dx={\int }_a^{b}F(a+b-x)dx]$
$\Rightarrow$ ${\int }_{-\pi }^{\pi }[\cos px+\sin qx{]}^2]dx=I$
$\Rightarrow$ $2{\int }_{-\pi }^{\pi }(\cos p^{2}x+\sin q^{2}x)dx=2I$
$\Rightarrow$ ${\int }_{-\pi }^{\pi }(\frac{(\cos 2px+1)}{2}+\frac{(\cos 2qx+1)}{2})dx=I$ since , $[{\cos }^{2}x=\frac{\cos 2x+1}{2}]$
$\Rightarrow$ $I=[\frac{\sin 2p\left(\pi \right)}{4p}+\frac{\sin 2q\left(\pi \right)}{4q}+\pi ]-[-\frac{\sin 2p\left(\pi \right)}{4p}-\frac{\sin 2q\left(\pi \right)}{4q}-\pi ]$
Since $[\sin k\pi =0]$ for any integral value of k. So,
$I=[0+0+\pi ]-[0+0-\pi ]$
Thus, $I=2\pi$
Hence, option 'D' is correct.