Integrals — Class 12 Maths Solution

Let $I = \int\limits_0^{\pi /4} {\cfrac{{\sin x\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}dx}$

Dividing numerator and denominator by ${\cos ^4}x$

,we get
$I = \int\limits_0^{\pi /4} {\cfrac{{\tan x{{\sec }^2}xdx}}{{1 + {{\tan }^4}x}}}$

Put ${\tan ^4}x = t$ $\Rightarrow$ $2\tan x{\sec ^2}xdx = dt$

When $x = 0,t = 0$ and when $x = \cfrac{\pi }{4},t = 1$

$\therefore$ $I = \cfrac{1}{2}\int\limits_0^1 {\cfrac{{dt}}{{1 + {t^2}}} = \left[ {\cfrac{1}{2}{{\tan }^{ - 1}}t} \right]_0^1} = \cfrac{1}{2} \times \cfrac{\pi }{4} = \cfrac{\pi }{8}$