Question

Evaluate: \[\int\limits_0^{\pi/2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x}dx\] .

Answer 1

\[Let I = \int_0^\frac{\pi}{2} \frac{x\sin x \cos x}{\sin^4 x + \cos^4 x}dx . \]

\[\text{ Then we have }: \]

\[I = \int_0^\frac{\pi}{2} \frac{\left( \frac{\pi}{2} - x \right)\sin\left( \frac{\pi}{2} - x \right) \cos\left( \frac{\pi}{2} - x \right)}{\sin^4 \left( \frac{\pi}{2} - x \right) + \cos^4 \left( \frac{\pi}{2} - x \right)}dx\]

\[\Rightarrow I = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}dx - \int_0^\frac{\pi}{2} \frac{x\sin x \cos x}{\sin^4 x + \cos^4 x} dx\]

\[\Rightarrow I = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}dx - I\]

\[\Rightarrow 2I = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x}dx\]

 Dividing the numerator and the denominator of RHS by cos⁴x, we have:

\[2I = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\tan x se c^2 x}{1 + \tan^4 x} dx\]

\[\Rightarrow 2I = \frac{\pi}{4} \int_0^\frac{\pi}{2} \frac{2\tan x se c^2 x}{1 + \tan^4 x} dx\]

\[\Rightarrow 2I = \frac{\pi}{4} \int_0^\frac{\pi}{2} \frac{2\tan x se c^2 x}{1 + \left( \tan^2 x \right)^2} dx\]

\[\text { Put} t = \tan^2 x\]

\[ \Rightarrow dt = 2\tan x se c^2 x dx\]

\[\text { When } x \to 0, t \to 0\]

\[\text { When } x \to \frac{\pi}{2}, t \to \infty\]

\[\therefore 2I = \frac{\pi}{4} \int_0^\infty \frac{1}{1 + t^2} dt\]

\[\Rightarrow 2I = \frac{\pi}{4} \left[ \tan^{- 1} \left( t \right) \right]_0^\infty \]

\[ \Rightarrow 2I = \frac{\pi}{4}\left[ \tan^{- 1} \left( \infty \right) - \tan^{- 1} \left( 0 \right) \right]\]

\[ \Rightarrow 2I = \frac{\pi}{4}\left[ \frac{\pi}{2} \right] = \frac{\pi^2}{8}\]

\[ \Rightarrow I = \frac{\pi^2}{16}\]