Question

solve $\underset{0}{\overset{11/2}{\int }}\frac{{\sin }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}dx$

Answer 1

$={\int }_0^{\frac{11}{2}}\frac{{\sin }^{2}x}{{\sin }^{2}x+{\cos }^{2}x}dx$

here , ${\sin }^{2}x+{\cos }^{2}x=1$

$={\int }_0^{\frac{11}{2}}{\sin }^{2}xdx$

we know , $2{\sin }^{2}x=1-\cos 2x$

$={\int }_0^{\frac{11}{2}}\frac{1-\cos 2x}{2}dx$

$={\int }_0^{\frac{11}{2}}\left(\frac{1}{2}+\frac{-\cos 2x}{2}\right)dx$

now, integrate

$={[\frac{x}{2}-\frac{\sin 2x}{4}]}_0^{\frac{11}{2}}$

$=(\frac{11}{4}-\frac{\sin 11}{4})-(0-0)$

$=\frac{11-\sin 11}{4}$