Question

Evaluate: $\stackrel{\pi }{\underset{0}{\int }}\frac{4x\sin x}{1+{\cos }^{2}x}dx$.

Answer 1

Let $I=$ $\stackrel{\pi }{\underset{0}{\int }}\frac{4x\sin x}{1+{\cos }^{2}x}dx$

or, $I=$ $\stackrel{\pi }{\underset{0}{\int }}\frac{4(\pi -x)\sin (\pi -x)}{1+{\cos }^2(\pi -x)x}dx$ [ Using property of definite integral]

or, $I=$ $\stackrel{\pi }{\underset{0}{\int }}\frac{4\pi \sin x}{1+{\cos }^{2}x}dx-I$

or, $2I=$ $(-4\pi )\stackrel{\pi }{\underset{0}{\int }}\frac{d\left(\cos x\right)}{1+{\cos }^{2}x}$

or, $2I=$ $(-8\pi )\stackrel{\frac{\pi }{2}}{\underset{0}{\int }}\frac{d\left(\cos x\right)}{1+{\cos }^{2}x}$ [ Using property of definite integral]

or, $I=(-4\pi ){\left[{\tan }^{-1}\cos x\right]}_0^{\frac{\pi }{2}}$

or, $I={\pi }^2$.