<s>[INST] Evaluating $\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$ The question is to evaluate  $$\int_0^{\pi/4} (\cos 2x)^{3/2} \cos x \, dx$$ I tried replacing $x$ by $\pi/4 -x$ and solving but couldn't get the answer.please help me in this regard.thanks.
 [/INST] $$
\cos(2x) = 1 - 2\sin^2 x.
$$
\begin{align}
\int_0^{\pi/4} (\cos(2x))^{3/2} \cos x\,dx & = \int_0^{\pi/4} \left( 1-2\sin^2 x \right)^{3/2} \overbrace{\Bigg( \cos x\,dx\Bigg)}^{\large\text{This will be } du.} \\[10pt]
& = \int_0^{1/\sqrt 2} (1-2u^2)^{3/2} \, du \\[10pt]
& = \int_0^1 (1-w^2)^{3/2} \, \frac{dw}{\sqrt 2} \\[10pt]
& = \frac 1 {2\sqrt 2} \int_{-1}^1 (1-w^2)^{3/2} \, dw \text{ (since this is an even function)} \\[10pt]
& = \frac 1 {2\sqrt 2} \int_{-1}^1 (1+w)^{3/2}(1-w)^{3/2} \, dw \\[10pt]
& = \frac 1 {2\sqrt 2} \int_0^1 (2v)^{3/2}(2(1-v))^{3/2} (2 \, dv) \\[10pt]
& = \frac 8 {\sqrt 2} \int_0^1 v^{3/2} (1 - v)^{3/2} \, dv \\[10pt]
& = \frac 8 {\sqrt 2} \operatorname{B}\left( \frac 5 2, \frac 5 2 \right) = \frac 8 {\sqrt 2} \cdot\frac{\Gamma(5/2)\Gamma(5/2)}{\Gamma(5)} = \frac {3\pi} {16\sqrt 2}.
\end{align}
APPENDIX:
\begin{align}
\int_0^1 x^{\alpha-1}(1-x)^{\beta-1} \, dx & = \operatorname{B}(\alpha,\beta) \quad \text{(the Beta function)} \\[10pt]
& = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)} \\[10pt]
\text{and } \Gamma(\alpha+1) & = \alpha\Gamma(\alpha), \\[10pt]
\text{so that } \Gamma\left( \frac 5 2 \right) & = \frac 3 2 \Gamma\left( \frac 3 2 \right) = \frac 3 2 \cdot \frac 1 2 \cdot \Gamma\left( \frac 1 2 \right) = \frac 3 2 \cdot \frac 1 2 \cdot \sqrt \pi.
\end{align}
</s>