Evaluate $ \int_{0}^{\pi/4}\left(\cos 2x \right)^{11/2}\cdot \cos x\;dx $
Applying the substitution $$s = \sqrt{2} \sin x, \qquad ds = \sqrt{2} \cos x \,dx $$ (which up to a constant is the one you suggest), we get $$\require{cancel} \int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{1}{\sqrt{2}} \int_0^1 (1 - s^2)^{11 / 2} ds.$$
Using integration by parts, we can rewrite the integral in $u$ in terms of an otherwise identical integral with a smaller exponent: For a general exponent $\alpha \neq 0$, taking $$u = (1 - s^2)^{\alpha}, \qquad dv = ds$$ gives $$\color{#00af00}{\int} \underbrace{\color{#00af00}{(1 - s^2)^{\alpha}}}_u \underbrace{\color{#00af00}{ds}}_{dv} = \underbrace{(1 - 2s^2)^{\alpha}}_u \underbrace{s}_v - \int \underbrace{s}_v \cdot \underbrace{\alpha (1 - s^2)^{\alpha - 1} \cdot (-2s) \,ds}_{du}.$$ Some (only slightly clever) manipulation of the integral on the r.h.s. gives $$\color{#00af00}{\int (1 - s^2)^{\alpha} ds} = s (1 - s^2)^{\alpha} - 2 \alpha \color{#00af00}{\int (1 - s^2)^{\alpha} ds} + 2 \alpha \int (1 - s^2)^{\alpha - 1} ds ,$$ and solving for our integral gives a reduction formula: $$\boxed{\color{#00af00}{\int (1 - s^2)^{\alpha} ds} = \frac{1}{2 \alpha + 1} s (1 - s^2)^{\alpha} + \frac{2 \alpha}{2 \alpha + 1} \int (1 - s^2)^{\alpha - 1} ds }.$$
If we start with a nonintegral half-integer $\frac{2 m - 1}{2}$, inductively applying this formula $m$ times yields expression for the antiderivative where the only integral expression that occurs is the familiar $$\int (1 - s^2)^{-1/2} ds = \arcsin s + C.$$ In our case, though, we need only the given definite integral, and our expression simplifies in a nice way when we specialize to our limits: $$ \int_0^1 (1 - s^2)^{\alpha} ds = \cancelto{0}{\left.\frac{1}{2 \alpha + 1} s (1 - s^2)^{\alpha}\right\vert_0^1} + \frac{2 \alpha}{2 \alpha + 1} \int_0^1 (1 - s^2)^{\alpha - 1} ds , $$ or a little more readably, $$ \phantom{(\ast)} \qquad \int_0^1 (1 - s^2)^{\alpha} ds = \frac{2 \alpha}{2 \alpha + 1} \int_0^1 (1 - s^2)^{\alpha - 1} ds. \qquad (\ast)$$
Taking $\alpha = \frac{11}{2}$ in $(\ast)$ gives $$\int_0^1 (1 - s^2)^{11 / 2} dt = \frac{11}{12} \int_0^1 (1 - s^2)^{9 / 2} ds,$$ and the integral on the r.h.s. is just the integral on the l.h.s. of the reduction formula with $\alpha = \frac{9}{2}$. Proceeding inductively thus gives $$\int_0^1 (1 - s^2)^{11 / 2} dt = \frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \int_0^1 (1 - s^2)^{-1 / 2} ds.$$ The integral on the r.h.s. is $$\int_0^1 (1 - s^2)^{-1 / 2} ds = \left.\arcsin s \right\vert_0^1 = \frac{\pi}{2} .$$ (Alternatively, stopping one step earlier gives the integral $\int_0^{1} \sqrt{1 - s^2} \,ds$, but this is just one-fourth the area of a unit circle, or $\frac{\pi}{4}$.) Now, putting everything together (and remembering the factor of $\frac{1}{\sqrt{2}}$ introduced by a change of variable to $s$) gives $$\color{#df0000}{\boxed{\int_0^{\pi / 4} (\cos 2x)^{11 / 2} \cos x \,dx = \frac{1}{\sqrt{2}} \cdot \frac{11}{12} \cdot \frac{9}{10} \cdot \frac{7}{8} \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} \cdot \frac{\pi}{2} = \frac{231 \pi}{2048 \sqrt{2}}}}.$$
Remark An induction argument along the same lines gives the general result $$\int_0^1 (1 - s^2)^{(2m - 1) / 2} ds = \frac{1}{4^m}{{2m}\choose{m}} \cdot \frac{\pi}{2} .$$
Interestingly, $$\frac{1}{4^m}{{2m}\choose{m}}$$ is both
- the coefficient of $r^m$ in the Maclaurin series for $\frac{1}{\sqrt{1 - r^2}}$,
- the probability of getting heads exactly half of the time when flipping a coin $2 m$ times,
either of which may well hint toward a slicker way of handling this family of integrals.
Use successive integrations by parts. Here is the first:
$$\int (1-2t^2)^{11/2} dt=t(1-2t^2)^{11/2}-\int \frac{11}{2}t (-4t)(1-2t^2)^{9/2}dt$$ $$=t(1-2t^2)^{11/2}-11\int (1-2t^2-1)(1-2t^2)^{9/2}dt$$ $$=t(1-2t^2)^{11/2}-11\int (1-2t^2)^{11/2}dt+11\int (1-2t^2)^{9/2}dt$$
Hence
$$12\int (1-2t^2)^{11/2} dt=t(1-2t^2)^{11/2}+11\int (1-2t^2)^{9/2}dt$$
When you have lowered the exponent enough, it's easy.
The following integrations yield
$$\int (1-2t^2)^{11/2} dt=\frac{1}{12}t(1-2t^2)^{11/2}+\frac{11}{12}\int (1-2t^2)^{9/2}dt$$
$$\int (1-2t^2)^{9/2} dt=\frac{1}{10}t(1-2t^2)^{9/2}+\frac{9}{10}\int (1-2t^2)^{7/2}dt$$
$$\int (1-2t^2)^{7/2} dt=\frac{1}{8}t(1-2t^2)^{7/2}+\frac{7}{8}\int (1-2t^2)^{5/2}dt$$
$$\int (1-2t^2)^{5/2} dt=\frac{1}{6}t(1-2t^2)^{5/2}+\frac{5}{6}\int (1-2t^2)^{3/2}dt$$
$$\int (1-2t^2)^{3/2} dt=\frac{1}{4}t(1-2t^2)^{3/2}+\frac{3}{4}\int (1-2t^2)^{1/2}dt$$
$$\int (1-2t^2)^{1/2} dt=\frac{1}{2}t(1-2t^2)^{1/2}+\frac{1}{2}\int (1-2t^2)^{-1/2}dt$$
The last is
$$\int \frac{1}{\sqrt{1-2t^2}}dt=\frac{1}{\sqrt{2}}\arcsin (\sqrt2 t)+C$$
Finally, modulo typing mistakes
$$\int (1-2t^2)^{11/2} dt= \frac{1}{12}t(1-2t^2)^{11/2} +\frac{11\cdot1}{12\cdot10}t(1-2t^2)^{9/2} +\frac{11\cdot9\cdot1}{12\cdot10\cdot8}t(1-2t^2)^{7/2} +\frac{11\cdot9\cdot7\cdot1}{12\cdot10\cdot8\cdot6}t(1-2t^2)^{5/2} +\frac{11\cdot9\cdot7\cdot5\cdot1}{12\cdot10\cdot8\cdot6\cdot4}t(1-2t^2)^{3/2} +\frac{11\cdot9\cdot7\cdot5\cdot3\cdot1}{12\cdot10\cdot8\cdot6\cdot4\cdot2}t(1-2t^2)^{1/2} +\frac{11\cdot9\cdot7\cdot5\cdot3\cdot1}{12\cdot10\cdot8\cdot6\cdot4\cdot2}\frac{1}{\sqrt{2}}\arcsin (\sqrt2 t)+C $$
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 $\Gamma(\tfrac12)=\sqrt\pi$.