Mathematics

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


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

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


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

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

Let $$t=\sin^{2}x\Rightarrow dt=2 \ \sin x \ \cos x \ dx$$

also,
$$\sin^2 {\dfrac {\pi} {2}}=1=upper limit$$

$$\sin^2 {0}=0=lower limit$$

Replacing we get,
$$2I=\dfrac{\pi}{4}\displaystyle\int_{0}^{1}\dfrac{1}{(1-t^{2})+t^{2}}dt$$, where 

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

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


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Subjective Medium Published on 17th 09, 2020
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