Question

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

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

Answer 1

Answer: (D)

Let $I={\int }_{-\pi }^{\pi }{(\cos px-\sin qx)}^{2}dx$
$={\int }_{-\pi }^{\pi }({\cos }^{2}px+{\sin }^{2}qx-2\cos px\sin qx)dx$
$=2{\int }_0^{\pi }({\cos }^{2}px+{\sin }^{2}qx)dx$
$-{\int }_{-\pi }^{\pi }2\cos px\sin qxdx$
$=2{\int }_0^{\pi }({\cos }^{2}px+{\sin }^{2}qx)dx-0$
$\left[\because {\int }_{-\pi }^{\pi }2\cos px\sin qxdxisanoddfunction\right]$
$=2{\int }_0^{\pi }\left[\frac{(1+\cos 2px)}{2}+\frac{1}{2}(1-\cos 2xqx)\right]dx$ $={\int }_0^{\pi }[1+\cos 2px+1-\cos 2qx]dx$
$={\int }_0^{\pi }[2+\cos 2px-\cos 2qx]dx$
$={\left[2x+\frac{\sin 2px}{2p}-\frac{\sin 2qx}{2p}\right]}_0^{\pi }=2\pi +0-0=2\pi$