Question

Evaluate the Integral.

Answer 1

\[\int\limits_{0}^{\pi/4}6\sec^4\theta \tan^4\theta\]

Answer 2

\[\int\limits_{0}^{\pi/4}6\sec^2 \theta \tan^4 \theta \sec^2 \theta d \theta\]substitute tan \(\theta=u\) then \[sec^2 \theta d \theta=du\] so \[\int\limits_{0}^{\pi /4}6(\tan^2 \theta+1)\tan^4 \theta \sec^2 \theta d \theta\]\[\int\limits_{0}^{\pi/4}6(u^2+1)u^4du\]