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.

Set $\cos 2x = \sin^2\theta$. Then
$$ \cos x = \sqrt{\frac{1+\sin^2\theta}{2}} \qquad \text{and} \qquad \left|\frac{dx}{d\theta}\right| = \frac{\sin\theta}{\sqrt{1+\sin^2\theta}}. $$
So we have
$$ \int_{0}^{\frac{\pi}{4}} \cos^{3/2}(2x) \cos x \, dx
= \frac{1}{\sqrt{2}} \int_{0}^{\frac{\pi}{2}} \sin^4\theta \, d\theta = \frac{3\pi}{16\sqrt{2}}. $$
Of course, you may recognize that this is simply a shortened version of previous answers.

Alternative solution. Apply the substitution $x = \theta/2$ and exploit the symmetry to write
$$ I
= \frac{1}{4} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cos\theta)^{3/2} e^{i\theta/2} \, d\theta
= \frac{1}{8\sqrt{2}} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (1 + e^{2i\theta})^{3/2} e^{-i\theta} \, d\theta. $$
Now we utilize the substitution $e^{i\theta} = iz$. If $\Gamma$ denotes the lower semicircular arc from $-1$ to $1$, the resulting integral is
$$ I = -\frac{1}{8\sqrt{2}} \int_{\Gamma} \frac{(1 - z^2)^{3/2}}{z^2} \, dz. $$
This integrand has the following antiderivative on the punctuated unit disc $\mathbb{D}\setminus\{0\}$:
$$ \frac{d}{dz} \left[ \left( \frac{1}{z} + \frac{z}{2} \right) \sqrt{1-z^2} + \frac{3}{2}\arcsin z \right] = - \frac{(1 - z^2)^{3/2}}{z^2}. $$
(This expression can be easily obtained by applying integration by parts twice. Of course, this back-of-the-envelope computation is then justified since the resulting function is holomoprhic on $\mathbb{D}\setminus\{0\}$ and has the desired derivative.) So $I$ can be computed as
$$ I = \frac{1}{8\sqrt{2}} \left[ \left( \frac{1}{z} + \frac{z}{2} \right) \sqrt{1-z^2} + \frac{3}{2}\arcsin z \right]_{z=-1}^{z=1} = \frac{3\pi}{16\sqrt{2}}.$$
