Evaluate the definite integral _0^ 4 x x ^4 x + ^4 x dx

Question

Evaluate the definite integral $\int_0^{\frac{\pi }{4}} {\frac{{\sin x \cos x}}{{{{\cos }^4}x + {{\sin }^4}x}}} dx$

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Answer

$I = \int_0^{\pi /4} {\frac{{\sin x.\cos x}}{{{{\cos }^4}x + {{\sin }^4}x}} dx}$ Dividing Nr. and Dr. by $\cos^4x$
$ = \int_0^{\pi /4} {\frac{{\frac{{\sin x.\cos x}}{\cos^4x}}}{{\frac{{{{\cos }^4}x}}{{{{\cos }^4}x}} + \frac{{{{\sin }^4}x}}{{{{\cos }^4}x}}}}dx} $
$ = \int_0^{\pi /4} {\frac{{\tan x.{{\sec }^2}x}}{{1 + {{\tan }^4}x}}dx} $
$ = \int_0^{\pi /4} {\frac{{\tan x.{{\sec }^2}x}}{{1 + {{({{\tan }^2}x)}^2}}}dx} $
Put ${\tan ^2}x = t$
$2\tan x.{\sec ^2}x\ d\ x = dt$
When $x = 0, t= 0$ and when $x=\frac{\pi }{4},t=1$
$\therefore{I} = \frac{1}{2}\int_0^1 {\frac{{dt}}{{1 + {t^2}}}} $
$ = \frac{1}{2}\left[ {{{\tan }^{ - 1}}t} \right]_0^1$
$ = \frac{1}{2}.\frac{\pi }{4} = \frac{\pi }{8}$

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