Evaluate $ \int_{0}^{\pi/4}\left(\cos 2x \right)^{11/2}\cdot \cos x\;dx $ 
Evaluate the definite integral
$$
I=\int_{0}^{\pi/4}(\cos 2x)^{11/2}\cdot \cos x\;dx
$$

My Attempt:
$$
I = \int \left(1-2\sin^2 x\right)^{11/2}\cdot \cos x\;dx
$$
Now, substitute $\sin x=t$ with $\cos x \,dx = dt$:
$$
I = \int (1-2t^2)^{11/2}\;dt
$$
How can I complete the solution from this point?
 A: Indeed, this special functions approach may be generalized.
Set $$j(s)=\int_0^{\pi/4}\cos(2x)^s\cos(x)dx.$$
Use $t=2\sin(x)^2$ so that $dt=4\sin(x)\cos(x)dx$, i.e. $\cos(x)dx=(2t)^{-1/2}dt$. Hence we have
$$j(s)=\frac1{\sqrt2}\int_0^1t^{1/2-1}(1-t)^{s+1-1}dt=\sqrt{\frac\pi2}\frac{\Gamma(s+1)}{\Gamma(s+\frac32)},$$
as $$\int_0^1t^{a-1}(1-t)^{b-1}dt=\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}$$
and $\Gamma(\tfrac12)=\sqrt\pi$.
A: Here's an efficient method using special functions.
Applying the substitution
$$t = 2 \sin^2 x, \qquad ds = 4 \sin x \cos x \,dx $$
gives
$$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt .$$
By the definition of and then a standard identity for the Beta function $\textrm{B}$,
$$\int_0^1 t^{-1/2} (1 - t)^{(2 m - 1) / 2} dt = \textrm{B} \left(\tfrac{1}{2}, m + \tfrac{1}{2}\right) = \frac{\Gamma(\tfrac{1}{2}) \Gamma(m + \frac{1}{2})}{\Gamma(m + 1)} .$$
where $\Gamma$ is the Gamma function, but we can easily handle all three of the factors $\Gamma(\,\cdot\,)$:


*

*By definition, $\Gamma\left(\frac{1}{2}\right) = \int_0^{\infty} q^{-1/2} e^{-q} \,dq ,$ and the substitution $q = r^2, dq = 2 r \, dr$ gives (by symmetry, one-half of) a Gaussian integral, yielding $$\Gamma\left(\tfrac{1}{2}\right) = \sqrt{\pi} .$$

*Applying the identity $\Gamma(z) = z \Gamma(z - 1)$ a total of $m$ times gives $$\Gamma\left(m + \tfrac{1}{2}\right) = \frac{(2 m)!}{4^m m!} \Gamma\left(\tfrac{1}{2}\right) = \frac{(2 m)!}{4^m m!} \sqrt\pi .$$

*Since $m$ is an integer, $\Gamma(m + 1) = m !$.


Assembling these pieces (and remembering the factor of $\frac{1}{2 \sqrt{2}}$ in the second display equation) gives
$$\int_0^{\pi / 4} (\cos 2x)^{(2 m - 1) / 2} \cos x \,dx = \frac{1}{2 \sqrt{2}} \frac{\left(\sqrt\pi\right)\left(\tfrac{(2 m)!}{4^m m!} \sqrt\pi\right)}{m!} = \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{4^m} {{2 m} \choose m} .$$
In our case, $\frac{2 m - 1}{2} = \frac{11}{2}$, so $m = 6$, and
$$\color{#bf0000}{\boxed{\int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{\pi}{2 \sqrt{2}} \cdot \frac{1}{4^6} {{12} \choose 6} = \frac{231 \pi}{2048 \sqrt{2}}}}.$$
A: I would suggest a different change of variables - 
Let $t=\cos(2x)$, so $dt=-4\cos(x)\sin(x)dx$ and we have $I=\int_0^{1} t^{11/2}\cdot (dt/4\sin(x))dt$
since $t=\cos(2x)=1-2\sin(x)^2$ we have $\sin(x)=\sqrt{(1-t)/2}$, and the integral reduces to:
$I=(\sqrt2/4)\int_0^{1} t^{11/2}(1-t)^{-1/2}dt$ which in my opinion is easier with integration by parts, comparing to the integral you already have.
