Question

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

Answer 1

\[Let I = \int_0^\frac{\pi}{2} \frac{x \sin x \cos x}{\sin^4 x + \cos^4 x} d x . . . (i)\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( \frac{\pi}{2} - x \right) \sin\left( \frac{\pi}{2} - x \right) \cos\left( \frac{\pi}{2} - x \right)}{\sin^4 \left( \frac{\pi}{2} - x \right) + \cos^4 \left( \frac{\pi}{2} - x \right)} d x\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( \frac{\pi}{2} - x \right)\cos x \sin x}{\cos^4 x + \sin^4 x} dx\]
\[ = \int_0^\frac{\pi}{2} \frac{\left( \frac{\pi}{2} - x \right)\sin x \cos x}{\sin^4 x + \cos^4 x} dx . . . (ii)\]
\[\text{Adding (i) and (ii) we get}\]
\[2I = \int_0^\frac{\pi}{2} \left( x + \frac{\pi}{2} - x \right)\frac{\sin x \cos nx}{\sin^4 x + \cos^4 x} d x\]
\[ = \frac{\pi}{2} \int_0^\frac{\pi}{2} \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} d x\]
\[\text{Let} \sin^2 x = t, \text{Then 2 sin x cosx}\ dx = dt\]
\[\text{When} x = 0, t = 0, x = \frac{\pi}{2}, t = 1\]
Therefore
\[2I = \frac{\pi}{4} \int_0^1 \frac{dt}{t^2 + \left( 1 - t \right)^2}\]
\[ = \frac{\pi}{8} \int_0^1 \frac{dt}{\left( t - \frac{1}{2} \right)^2 + \frac{1}{4}}\]
\[ = \frac{\pi}{8} \times 2 \left[ ta n^{- 1} \left( 2t - 1 \right) \right]_0^1 \]
\[ = \frac{\pi}{4}\left( \frac{\pi}{4} + \frac{\pi}{4} \right)\]
\[Hence\ I = \frac{\pi^2}{16}\]