Question

If ${\int }_0^{\pi }\frac{x^2}{(1+\sin x{)}^2}dx=A$ the ${\int }_0^{\pi }\frac{2x^{2}co{s}^2\left(x/2\right)}{(1+\sin x{)}^2}dx=$ ?

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

Answer 1

The correct option is A $A+\pi -{\pi }^2$

${\int }_0^{\pi }\frac{2x^{2}co{s}^2\left(x/2\right)}{(1+sinx{)}^2}dx={\int }_0^{\pi }\frac{x^2(1+cosx)}{(1+sinx{)}^2}dx$

$={\int }_0^{\pi }\frac{x^2}{(1+sinx{)}^2}dx+{\int }_0^{\pi }\frac{cosx}{(1+sinx{)}^2}dx$

On solving ${\int }_0^{\pi }\frac{x^{2}cosx}{(1+sinx{)}^2}dx$ through by parts we get;

${\int }_0^{\pi }\frac{2x^{2}co{s}^2\left(x/2\right)}{(1+sinx{)}^2}dx=A+\pi -{\pi }^2$