Find $$\int_0^{\frac{\pi}{4}}(\cos2x)^{3/2}\cos x dx$$
My attempt:
$$I=\int_0^{\frac{\pi}{4}}(\cos2x)^{3/2}\cos x dx=\int_0^{\frac{\pi}{4}}(1-2\sin^2x)^{3/2}\cos x dx$$
Let $\sin x=t$
$$I=\int_0^{\frac{1}{\sqrt2}}(1-2t^2)^{3/2}dt$$
I am stuck here.
Prologue
I won't fix the extrema of the integral after my substitution, and I'll leave you the renaming part because it's simple
HINT
From where you got stuck, substitute
$$t = \frac{1}{\sqrt{2}}\sin(y) ~~~~~~~~~~~ \text{d}t = \frac{1}{\sqrt{2}}\cos(y)\ \text{d}y$$
So
$$(1 - 2t^2)^{3/2} = (1 - \sin^2(y))^{3/2} = \cos^3(y)$$
Remembering now that
$$y = \arcsin(\sqrt{2}t)$$
You have to integrate
$$\frac{1}{\sqrt{2}}\int\cos^4(y)\ \text{d}y$$
Useful Reduction Formula
https://en.wikipedia.org/wiki/Integration_by_reduction_formulae
You will fine the Cosine reduction formula in the page!
