Question

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

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

Answer 1

Answer: (D)

$2\pi$

${\int }_{-\pi }^{\pi }(\cos px-\sin qx{)}^{2}dx$

$={\int }_{-\pi }^{\pi }(\sin qx-\cos px{)}^{2}dx$... Since the given function is even

$={\int }_{-\pi }^{\pi }({\sin }^{2}qx+{\cos }^{2}px-2\cos px\sin qx)dx$

$=\left[{\int }_{\pi }^{\pi }\left({\sin }^{2}qx\right)dx+{\int }_{-\pi }^{\pi }\left({\cos }^{2}px\right)dx-{\int }_{-\pi }^{\pi }\left(2\cos px\sin qx\right)dx\right]$

Consider, $\int {\sin }^{2}qxdx$

Put $qx=u⟹dx=\frac{du}{q}$

$\int {\sin }^{2}qxdx=\frac{1}{q}\int {\sin }^{2}udu$

$=\frac{1}{q}\int \frac{1}{2}du-\frac{\cos u\sin u}{2q}$

$=\frac{u}{2q}-\frac{\sin \left(2u\right)}{4q}$

$=\frac{qx}{2q}-\frac{sin\left(2qx\right)}{4q}$

Similarly,

$\int {\cos }^{2}pxdx=\frac{1}{p}\int {\cos }^{2}udu$

$=\frac{1}{p}\int \frac{1}{2}du-\frac{\cos u\sin u}{2p}$

$=\frac{u}{2p}-\frac{\sin \left(2u\right)}{4p}$

$=\frac{x}{2}-\frac{\frac{\sin \left(2px\right)}{2}}{2p}$

Consider, $\int 2\cos px\sin qxdx$

$=\int \sin (p+q)x+\sin (q-p)xdx$

$=-\frac{\cos \left(\right(p+q\left)x\right)}{(q+p)}-\frac{\cos \left(\right(q-p\left)x\right)}{(q-p)}$

Then, ${\int }_{-\pi }^{\pi }(\cos px-\sin qx{)}^{2}dx$

$={|\frac{x}{2}-\frac{\frac{\sin \left(2qx\right)}{2}}{2q}|}_{-\pi }^{\pi }$$+{|\frac{x}{2}-\frac{\frac{\sin \left(2px\right)}{2}}{2p}|}_{-\pi }^{\pi }$$+{|-\frac{\cos \left(\right(p+q\left)x\right)}{(q+p)}-\frac{\cos \left(\right(q-p\left)x\right)}{(q-p)}|}_{-\pi }^{\pi }$

$=\pi +\pi =2\pi$