$\int_0^{\frac{\pi}{4}} \sec ^4 x d x=$

Question

Aicharya · Accepted Answer

$\begin{aligned} & \int_0^{\frac{\pi}{4}} \sec ^4 x \mathrm{~d} x=\int_0^{\frac{\pi}{4}}\left(	an ^2 x+1ight) \sec ^2 x \mathrm{~d} x \ & =\left[\frac{	an ^3 x}{3}+	an xight]_0^{\frac{\pi}{4}} \ & =\frac{1}{3}+1 \ & =\frac{4}{3}\end{aligned}$

$\int_0^{\frac{\pi}{4}} \sec ^4 x d x=$

Solution