calculus - How can one show that $int_{0}^{pi/4}{sqrt{sin(2x)}over cos^2(x)}mathrm dx=2-sqrt{2over pi}cdotGamma^2left({3over 4}right)?$

Proposed:

$$\int_{0}^{\pi/4}{\sqrt{\sin(2x)}\over \cos^2(x)}\mathrm dx=2-\sqrt{2\over \pi}\cdot\Gamma^2\left({3\over 4}\right)\tag1$$

My try:

Change $(1)$ to

$$\int_{0}^{\pi/4}\sqrt{2\sec^2(x)\tan(x)}\mathrm dx\tag2$$

$$\int_{0}^{\pi/4}\sqrt{2\tan(x)+2\tan^3(x)}\mathrm dx\tag3$$

Not sure what substitution to use

How may we prove $(1)?$

Answer

By substituting $x=\arctan t$ our integral takes the form:

$$I=\int_{0}^{1}\sqrt{\frac{2t}{1+t^2}}\,dt$$

and by substituting $\frac{2t}{1+t^2}=u$ we get:
$$I = \int_{0}^{1}\left(-1+\frac{1}{\sqrt{1-u^2}}\right)\frac{du}{u^{3/2}}$$
that is straightforward to compute through the substitution $u^2=s$ and Euler's Beta function:
$$I = \frac{1}{2} \left(4+\frac{\sqrt{\pi }\,\Gamma\left(-\frac{1}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}\right).$$
The identities $\Gamma(z+1)=z\,\Gamma(z)$ and $\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin(\pi z)}$ settle OP's $(1)$.