7.2 definite integral — Maths STD 12 Science — Question

$I = \int_0^{\pi /8} {\,(1 - {{\sin }^2}4\theta )\cos 4\theta \,d\theta } $

Put $\sin 4\theta = t \Rightarrow \cos 4\theta \,d\theta = \frac{{dt}}{4}$

When $\theta = 0 \to \frac{\pi }{8},$ then $t = 0 \to 1$

$\therefore$  $I = \frac{1}{4}\int_0^1 {(1 - {t^2})dt = \frac{1}{4}} \left[ {t - \frac{{{t^3}}}{3}} \right]_0^1 = \frac{1}{6}$.