Question

If ${\int }_0^{\pi }x{\cos }^2\left(\sin x\right)dx=a$ then find value of ${\int }_0^{\pi }{\sin }^2\left(\sin x\right)dx?$

Answer 1

Given : ${\int }_0^{\pi }x{\cos }^2\left(\sin x\right)dx=a$

$I={\int }_0^{\pi }x{\cos }^2\left(\sin x\right)dx=a.........\left(1\right)I={\int }_0^{\pi }(\pi -x){\cos }^2\left(\sin (\pi -x)\right)dxI={\int }_0^{\pi }(\pi -x){\cos }^2\left(\sin \left(x\right)\right)dx........\left(2\right)$

Adding (1) and (2)

$2I={\int }_0^{\pi }\pi {\cos }^2\left(\sin \left(x\right)\right)dx2I=\pi {\int }_0^{\pi }(1-{\sin }^2\left(\sin x\right))dx2I=\pi [{\int }_0^{\pi }dx-{\int }_0^{\pi }{\sin }^2\left(\sin x\right)dx]2I=\pi (\pi -0)-{\int }_0^{\pi }{\sin }^2\left(\sin x\right)dx$

From equation $1$

$2a={\pi }^2-{\int }_0^{\pi }{\sin }^2\left(\sin x\right)dx{\int }_0^{\pi }{\sin }^2\left(\sin x\right)dx={\pi }^2-2a$

Hence the correct answer is ${\pi }^2-2a$