Mathematics

Evaluate the following integral :
$$\displaystyle\int_{0}^{\pi/2}\dfrac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\ dx$$


SOLUTION
We have, 
$$I=\displaystyle\int_{0}^{\pi/2}\dfrac{x\sin x\cos x}{\sin^{4}x+\cos^{4}x}\ dx$$  ....(i)

As we know

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


$$\Rightarrow I=\displaystyle\int_{0}^{\pi/2}\dfrac{\left(\dfrac{\pi}{2}-x\right)\cos x\sin x}{\cos^{4}x+\sin^{4}x}dx$$   .....(ii)

Adding $$(i)$$ and $$(ii)$$ , we get

$$2I=\dfrac{\pi}{2}\displaystyle\int_{0}^{\pi/2}\dfrac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}dx$$

$$\Rightarrow 2I=\dfrac{\pi}{4}\displaystyle\int_{0}^{1}\dfrac{1}{(1-t^{2})+t^{2}}dt$$,                   where $$t=\sin^{2}x\Rightarrow \dfrac{dt}{2}=sinx\,cosx\,dx$$

$$2I=\dfrac{\pi}{8}\dfrac{\pi}{8}\displaystyle\int_{0}^{1}\dfrac{1}{\left(t-\dfrac{1}{2}\right)^{2}+\left(\dfrac{1}{2}\right)^{2}}dt=\dfrac{\pi}{8}\times 2\left[\tan^{-1} (2t-1)\right]_{0}^{1}$$

$$\Rightarrow I=\dfrac{\pi}{8}\left(\dfrac{\pi}{4}+\dfrac{\pi}{4}\right)=\dfrac{\pi^{2}}{16}$$

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