Question

${\int }_0^{\pi /2}\frac{si{n}^{2}x}{(1+cosx{)}^2}dx=.....$

• A. $\frac{4-\pi }{2}$
• B. $\frac{\pi -4}{2}$
• C. $4-\frac{\pi }{2}$
• D. $\frac{4+\pi }{2}$

Answer 1

The correct option is A $\frac{4-\pi }{2}$

${\int }_0^{\pi /2}\frac{{\sin }^{2}x}{{(1+\cos x)}^2}={\int }_0^{\pi /2}\frac{1-{\cos }^{2}x}{{(1+\cos x)}^2}dx$

$={\int }_0^{\pi /2}\frac{(1+\cos x)(1-\cos x)}{{(1+\cos x)}^2}dx$

$={\int }_0^{\pi /2}\frac{1-\cos x}{1+\cos x}dx={\int }_0^{\pi /2}\frac{2{\sin }^{2}x/2}{2{\cos }^{2}x/2}dx$

$={\int }_0^{\pi /2}{\tan }^2\frac{x}{2}dx$

$={\int }_0^{\pi /2}({\sec }^{2}x/2-1)dx$

$=\frac{\tan x/2}{1/2}-x=2\tan \frac{x}{2}-x{]}_0^{\pi /2}$

$=2\tan \frac{\pi }{4}-\frac{\pi }{2}=2-\frac{\pi }{2}$$=\frac{4-\pi }{2}$