Mathematics

Prove:
$$\int _ { 0 } ^ { \pi } \dfrac { x \sin x } { 1 + \cos ^ { 2 } x } d x = \dfrac { \pi ^ { 2 } } { 4 }$$


SOLUTION
$$I = \int_{0}^{\pi} \dfrac {x \, sin \,x}{1+ cos^2x}dx$$...(1)
using property $$\infty $$ definite integral

$$\int_{a}^{b} f(x)dx =\int_{a}^{b} f(a+b-x)dx$$

$$I = \int_{a}^{b}\dfrac {(\pi-x)sin (\pi-x)}{1+cos^2(\pi-x)}dx$$

$$I = \int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}dx$$....(2)

adding $$(1)$$& $$(2)$$, we get

$$I+I=\int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}dx + \int_{a}^{b}\dfrac {(\pi-x)sinx}{1+cos^2x}$$

$$ 2I = \int_{a}^{b}\dfrac {(\pi- Sin x)sinx}{1+cos^2x}dx$$

$$I =\dfrac {\pi}{2}\int_{0}^{\pi} \dfrac {Sin x}{1+Cos^2x}dx$$

$$=\dfrac {\pi}{2}\cdot 2 \int_{0}^{\pi/2} \dfrac {Sin x}{1+Cos^2x}dx$$

$$= -\pi \int_{0}^{\pi/2} \frac {d(cos\,x)}{1+Cos^2x}dx$$

$$= - \pi [tan^{-1}(cos\,x)]_0^{\pi/2}$$

$$-\pi [ tan^{-1} cos \dfrac{\pi}2-tan^{-1}(cos \,0)]$$

$$-\pi [ 0- \dfrac{\pi} 4]$$

$$=\dfrac{\pi^2}4$$
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