Question

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

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

Answer 1

The correct option is D $2\pi$

$I={\int }_{-\pi }^{\pi }{(\cos ax-\sin bx)}^{2}dx$

$={\int }_{-\pi }^{\pi }({\cos }^{2}ax+{\sin }^{2}bx-2\cos ax.\sin bx)dx$

$={\int }_{-\pi }^{\pi }({\cos }^{2}ax+{\sin }^{2}bx)dx$, [ Since $\cos ax.\sin bx$ is odd function ].

$=2{\int }_0^{\pi }({\cos }^{2}ax+{\sin }^{2}bx)dx$

$={\int }_0^{\pi }\left[\right(1+\cos 2ax)+(1-\cos 2bx\left)\right]dx$
$=2.{\int }_0^{\pi }dx+{\int }_0^{\pi }(\cos 2ax-\cos 2bx)dx$
$=2\pi +{(\frac{\sin 2ax}{2a}-\frac{\sin 2bx}{2b})}_0^{\pi }=2\pi$
Since $a,b$ are integer.