Question

${\int }_{-\pi }^{\pi }{(\cos ax-\sin bx)}^{2}dx$ where a and b are integer is equal to

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

Answer 1

The correct option is D $2\pi$

${\int }_{-\pi }^{\pi }(\cos ax-\sin bx)dx={\int }_{-\pi }^{\pi }({\cos }^{2}ax+{\sin }^{2}bx-2\cos ax\sin bx)dx$
$={\int }_{-\pi }^{\pi }(\frac{{\cos }^{2}ax}{2}+\frac{1}{2})dx+{\int }_{-\pi }^{\pi }(\frac{\cos 2ax}{2}-\frac{1}{2})dx$
$-2{\int }_{-\pi }^{\pi }(\sin (2a+2b)x-\sin (2a+2b)x)dx$
$={[\frac{\sin 2ax}{4a}-\frac{2\sin bx}{4a}+x-\frac{\cos \left(x(a-b)\right)}{a-b}+\frac{\cos \left(x(a+b)\right)}{a+b}]}_{-\pi }^{\pi }=2\pi$