Mathematics

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


SOLUTION
Let $$I=\displaystyle\int^{\pi}_0\dfrac{x\sin x}{1+\cos^2x}dx$$ ………$$(1)$$

$$I=\displaystyle\int^{\pi}_0\dfrac{(\pi -x)\sin(\pi-x)}{1+\cos^2(\pi-x)}dx$$

$$=\displaystyle\int^{\pi}_0\dfrac{(\pi -x)\sin x}{1+\cos^2x}dx$$ ……..$$(2)$$

Adding both $$(1)$$ and $$(2)$$, we get

$$2I=\displaystyle\int^{\pi}_0\dfrac{\pi\sin x}{1+\cos^2x}dx$$

Let $$t=\cos x$$             $$\cos 0=1$$

$$dt=-\sin xdx$$              $$\cos\pi =-1$$

So, $$2I=\displaystyle\int^{-1}_1\dfrac{-\pi dt}{1+t^2}$$

$$2I=\pi\displaystyle\int^1_{-1}\dfrac{dt}{1+t^2}=\pi\left[\tan^{-1}t\right]^1_{-1}$$

$$2I=\pi\left[\dfrac{\pi}{4}-\left(-\dfrac{\pi}{4}\right)\right]$$

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

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

$$I=\left(\dfrac{\pi}{2}\right)^2$$.
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Subjective Medium Published on 17th 09, 2020
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