$$\int ^\frac{\pi}{2} _0 \cos^4(s)\mathrm{ds}-\int^\frac{\pi}{2}_0 \cos^6(s)\mathrm{ds} $$
the answer is $\dfrac{3\cdot1}{4\cdot2}-\dfrac{5\cdot3\cdot1}{6\cdot4\cdot2}$
I really don't know why
There may be some tricks in this, but I have not seen it in the textbook
Integrate-by-parts below to obtain the recursive formula
\begin{align} I_n=\int ^\frac{\pi}{2} _0 \cos^n s \>{ds}&= \int ^\frac{\pi}{2}_0 \frac{\cot^{n-1} s }{n}d(\sin^{n}s) {ds}=\frac{n-1}nI_{n-2},\>\>\>I_0=\frac\pi2 \end{align}
and then apply it to
$$\int^\frac{\pi}{2}_0 \cos^4s \> {ds} -\int^\frac{\pi}{2}_0 \cos^6s \>{ds} =I_4-I_6=\frac34\frac12I_0-\frac56 \frac34\frac12I_0$$
Define $$I_n=\int_0^{\pi/2}\cos^{2n}(t)dt$$
Integrate by parts $$I_n=\int_0^{\pi/2}\cos^{2n-1}(t)\cos(t)dt=\color{blue}{\cos^{2n-1}(t)\sin(t)\Big|_0^{\pi/2}}+(2n-1)\int_0^{\pi/2}\cos^{2n-2}(t)\sin^2(t)dt$$ and observe that the blue part is equal to zero.
Use $\cos^2(t)+\sin^2(t)=1$ $$(2n-1)\int_0^{\pi/2}\cos^{2n-2}(t)\sin^2(t)dt=(2n-1)\Big(\int_0^{\pi/2}\cos^{2(n-1)}(t)dt-\int_0^{\pi/2}\cos^{2n}(t)dt\Big)$$
Collect all results $$I_n=(2n-1)(I_{n-1}-I_n)$$
Isolate $I_n$ $$I_n=\frac{2n-1}{2n}I_{n-1}$$
Compute $I_0, I_1, I_2, I_3$ $$I_0=\frac{\pi}{2},\quad I_1=\frac{1}{2}\frac{\pi}{2},\quad I_2=\frac{3\cdot 1}{4\cdot 2}\frac{\pi}{2},\quad I_3=\frac{5\cdot 3\cdot 1}{6\cdot 4\cdot 2}\frac{\pi}{2}$$
Compute $I_2-I_3$
It seems that one of the solution may be the formula for $\cos$ at even power: $\cos^{2n}(s)=\frac{1}{2^{2n}}(e^{is}+e^{-is})^{2n}$: $I(n)=\int ^\frac{\pi}{2} _0 \cos^{2n}(s)\mathrm{ds}=\frac{1}{4}\int ^{2\pi} _0 \cos^{2n}(s)\mathrm{ds}=\frac{1}{2^{2n}4}\int ^{2\pi} _0 (e^{is}+e^{-is})^{2n}\mathrm{ds}=$$\frac{1}{2^{2n}4}\int ^{2\pi} _0 (e^{2ins}+\binom{2n}{1}e^{is(2n-1)-is}+...+\binom{2n}{n}e^{is(2n-n)-isn}+...+e^{-2ins})\mathrm{ds}$
After integration the only surviving term is $\frac{2\pi}{2^{2n}4}\binom{2n}{n}=\frac{2\pi(2n)!}{2^{2n}4(n!)^2}$
