Mathematics

The value of $$\int\limits_0^{\pi /2} {\dfrac{{x\sin x\cos x}}{{{{\sin }^4}x + {{\cos }^4}x}}dx\,is} $$ is 


ANSWER

$$\dfrac{{{\pi ^2}}}{{16}}$$


SOLUTION
$$ \displaystyle I = \int_{0}^{\pi/2} \frac{x sinx \, cosx}{sin^4x + cos^4 x}dx$$
using $$  \displaystyle  \int_{0}^{a} f(n)dx = \int_{0}^{a} f(a-x) dx, $$ we get 
$$  \displaystyle  I= \int_{0}^{\pi/2} \frac{(\pi/2 -\pi) sin (\pi /2 -\pi) cos (\pi/2 - \pi)}{sin^4 (\pi/2 -\pi)+ cos^4 (\pi /2 -\pi)}$$
$$  \displaystyle  = \int_{0}^{\pi/2} \frac{(\pi/2-\pi)cosx sinx}{cos^4x + sin^4 x} dx $$
$$  \displaystyle  I = \int_{0}^{\pi/2} \frac{cos sinx}{cos^4 x+ sin^4 x}dx - \int_{0}^{\pi/2} \frac{+xcosx \, sinx}{cos^4x+sin^4x}dx $$
$$  \displaystyle \therefore 2I = \frac{\pi}{2} \int_{0}^{\pi/2} \frac{sinx cosx}{sin^4 x + cos^4x} dx - I $$
$$  \displaystyle \therefore I = \dfrac{\pi}{2}\int_{0}^{1} \frac{dt/2}{t^2+(1-t)^2}$$
as $$ x = \pi / 2, $$ then t = 1
and $$ x = 0 , $$ then $$ t =0 $$
$$  \displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{t^2+t^2+1-2t}$$
$$  \displaystyle = \frac{\pi}{8} \int_{0}^{1} \dfrac{dt}{2t^2-2t+1}$$
$$  \displaystyle  = \dfrac{\pi}{16} \int_{0}^{1} \dfrac{dt}{(t-\dfrac{1}{2})^2 + (\dfrac{1}{2})^2}$$
$$  \displaystyle  = \dfrac{\pi}{16} \times \frac{1}{1/2} [ tan^{-1} (\frac{t-1/2}{1/2})]$$
$$  \displaystyle \because \int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1}x/a$$
$$  \displaystyle  =  [tan^{-1} 2(1-1/2)-tan^{-1}2(0-1/2)]$$
$$  \displaystyle  = \dfrac{\pi}{8} [tan^{-1}1-tan^{-1}(-1)]$$
$$  \displaystyle \because tan^{-1}(-1) = -tan^{-1}1 = -\pi/4$$
$$  \displaystyle  I = \frac{\pi}{8} [ \frac{\pi}{4}+\frac{\pi}{4}] = \frac{\pi^2}{16}$$

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Single Correct Medium Published on 17th 09, 2020
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