Integrals — Maths STD 12 Science — Question

Dividing Nr. and Dr. by cos⁴x

$ = \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 ${\tan ^2}x = t$

$2\tan x.{\sec ^2}xdx = dt$

When x = 0, t= 0 and when $x=\frac{π}{4},t=1$

$\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}$