Question

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

Answer 1

\[Let, I = \int_0^\frac{\pi}{2} \frac{x\sin x \cos x}{\sin^4 x + \cos^4 x} d x...............(1)\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( \frac{\pi}{2} - x \right)\sin\left( \frac{\pi}{2} - x \right) \cos\left( \frac{\pi}{2} - x \right)}{\sin^4 \left( \frac{\pi}{2} - x \right) + \cos^4 \left( \frac{\pi}{2} - x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( \frac{\pi}{2} - x \right)\cos x \sin x}{\cos^4 x + \sin^4 x}dx ..............(2)\]
Adding (1) and (2)
\[2I = \int_0^\frac{\pi}{2} \frac{\left( x + \frac{\pi}{2} - x \right)\sin x \cos x}{\sin^4 x + \cos^4 x} d x \]
\[ = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{\left( \sin^2 x + \cos^2 x \right)^2 - 2 \sin^2 x \cos^2 x} d x\]
\[ = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{1 - 2 \sin^2 x \cos^2 x}dx\]
\[ = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{1 - 2 \sin^2 x \left( 1 - \sin^2 x \right)}dx\]
\[ = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{1 - 2 \sin^2 x + 2 \sin^4 x}dx\]
\[\text{Let, }\sin^2 x = t,\text{ then }2\sin x\cos x dx = dt \]
\[\text{When, }x \to 0 ; t \to 0\text{ and }x \to \frac{\pi}{2} ; t \to 1\]
\[ 2I = \frac{\pi}{4} \int_0^1 \frac{1}{1 - 2t + 2 t^2}dt\]
\[ = \frac{\pi}{8} \int_0^1 \frac{1}{\left( t - \frac{1}{2} \right)^2 + \frac{1}{4}}\]
\[ = \frac{\pi}{8} \left[ 2 \tan^{- 1} \left( 2t - 1 \right) \right]_0^1 \]
\[ = \frac{\pi}{4}\left[ \tan^{- 1} \left( 1 \right) - \tan^{- 1} \left( - 1 \right) \right]\]
\[ = \frac{\pi}{4}\left[ \frac{\pi}{4} + \frac{\pi}{4} \right]\]
\[ = \frac{\pi^2}{8}\]
\[Hence, I = \frac{\pi^2}{16}\]