$\int_{-\pi / 4}^{\pi / 4} \cos ^{-8} x d x=$

Question

Aicharya · Accepted Answer

$ I=\int_{-\pi / 4}^{\pi / 4} \frac{1}{\cos ^8 x} d x=\int_{-\pi / 4}^{\pi / 4}\left(\sec ^6 xight)\left(\sec ^2 xight) d x $ On putting $	an x=t \Rightarrow \sec ^2 x d x=d t$ $ \begin{aligned} 	herefore I & =\int_{-1}^1\left(t^2+1ight)^3 d t=\int_{-1}^1\left(t^6+3 t^4+3 t^2+1ight) d t \ & =2 \int_0^1\left(t^6+3 t^4+3 t^2+1ight) d t \ & =2\left[\frac{t^7}{7}+\frac{3 t^5}{5}+t^3+tight]_0^1=2\left[\frac{1}{7}+\frac{3}{5}+2ight] \ & =\frac{192}{35} \end{aligned} $

$\int_{-\pi / 4}^{\pi / 4} \cos ^{-8} x d x=$

Solution