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Q.
$\int\limits_0^{\pi / 4}(\cos 2 x)^{3 / 2} \cdot \cos x d x=$
Integrals
Solution:
$ I =\int\limits_0^{\pi / 4}(\cos 2 x )^{3 / 2} \cos xdx $
$=\int\limits_0^{\pi / 4}\left(1-2 \sin ^2 x \right)^{3 / 2} \cos xdx $
$\text { put } \sqrt{2} \sin x = t \Rightarrow \cos x d x = dt / \sqrt{2}$
$ I =\frac{1}{\sqrt{2}} \int\limits_0^1\left(1- t ^2\right)^{3 / 2} dt$
Now let $t =\sin \theta \Rightarrow dt =\cos \theta d \theta$
$I=\frac{1}{\sqrt{2}} \int\limits_0^{\pi / 2} \cos ^4 \theta d \theta=\frac{3 \pi}{16 \sqrt{2}}$