Question

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

Answer 1

Let $I={\int }_{-x}^x(\cos ax-\sin bx{)}^{2}dx$
$={\int }_{-x}^x({\cos }^{2}ax+{\sin }^{2}bx-2\cos ax\sin bx)dx$--- (1)

Since ${\cos }^{a}x$ and ${\sin }^{2}bx$ are even function and $\cos ax\sin bx$ is an odd function, therefore,

${\int }_{-x}^x{\cos }^{2}axdx=2{\int }_0^x{\cos }^{2}axdx$

${\int }_{-x}^x{\sin }^{2}bxdx=2{\int }_0^x{\sin }^{2}bxdx$
and ${\int }_{-x}^x\cos ax\sin bxdx=0$

From (1), we get
$I=2{\int }_0^x{\cos }^{2}axdx+2{\int }_0^x{\sin }^{2}bxdx=2{\int }_0^x\frac{1+\cos 2ax}{2}dx+2{\int }_0^x\frac{1-\cos 2bx}{2}dx$

$2{\int }_0^{x}1dx+{\int }_0^x(\cos 2ax-\cos 2bx)dx=2[x{]}_0^x+{[\frac{\sin 2ax}{2a}-\frac{\sin 2bx}{2b}]}_0^x$

$I=2x+\frac{1}{2a}\sin 2ax-\frac{1}{2b}\sin 2bx$