Question

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

Answer 1

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

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

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

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

$\Rightarrow I=2{\int }_0^{\pi }({\cos }^{2}ax+{\sin }^{2}bx)dx-0$

[$\because ({\cos }^{2}ax+{\sin }^{2}bx)$ is an even function while $2\sin axcosbx$ is an odd function]

$\Rightarrow I={\int }_0^{\pi }{\cos }^{2}axdx+2{\int }_0^{\pi }{\sin }^{2}bxdx$

$\Rightarrow I=2{\int }_0^{\pi }\frac{(1+\cos 2ax)dx}{2}+{\int }_0^{\pi }\left(\frac{1-\cos 2bx}{2}\right)dx$

$\Rightarrow I={[x+\frac{\sin 2ax}{2a}]}_0^{\pi }+{[x-\frac{\sin 2bx}{2b}]}_0^{\pi }$

$\Rightarrow I=\pi +\frac{\sin 2a\pi }{2a}+\pi -\frac{\sin 2b\pi }{2b}$

$\Rightarrow I=2\pi +\frac{\sin 2a\pi }{2a}-\frac{\sin 2b\pi }{2b}$