Reduction formula for $\int \cos^n(x) dx$. Why this way and not any other way? I feel so overwhelmed!
for (cos(x))^n, everywhere I see the same thing:

But there are so many other possible routes, how on earth did they just know what do!?
Could have said:
$(cosx)^n=(cosx)^{n-2}(1-sinx^2)$
or $(cosx)^{n-3}(1-sinx^2)(cosx)$
andevery possible route has different branching routes! I feel so frustrated!
 A: There are many ways to decompose $\cos^n(x)$ as you say.
However it must be borne in mind that it's the subsequent integral of the resulting functions that needs to be computed. The chosen decomposition ideally leads to convenient further integration by parts generating further powers of $\cos$ alone i.e. $I_n$
If you chose e.g. $\cos^n(x)=\cos^{n-2}(x)(1-\sin^2(x))$ then you'd be left with 
$$
I_{n-2}-\int \cos^{n-2}(x)\sin^2(x)\,dx
$$
The second integral cannot now be done with the trig. substitution $\sin^2=1-\cos^2$ as that just takes you back to your starting point, so you need another way of doing it, and that may or may not be obvious to you:
$$
\int \cos^{n-2}(x)\sin^2(x)\,dx =\int \cos^{n-2}(x)\sin(x)\sin(x)\,dx
$$
This can be integrated by parts by setting
$dv=\cos^{n-2}(x)\sin(x)$ and $u=\sin(x)$
giving 
$$
-\sin(x)\frac{1}{n-1}\cos^{n-1}(x)+\frac{1}{n-1}\int\cos^{n-1}(x)\cos(x)\,dx
$$
and the last integral is $I_n$ again.
So, it is possible to do it other ways. I feel this second way though is less convenient and less direct than the way you described. I think other decompositions will also be less convenient though it may be a moot point what is more convenient or not. 
A: Yes you can re write as -
$$(\cos x)^n=(\cos x)^{(n-2)}(1-\sin^2x)$$
or $$(\cos x)^{(n-3)}(1-\sin^2x)(\cos x)$$
But then at some stage you need to change $$\sin^2x$$ into $$\frac{1-\cos 2x}{2}$$ That leads to you more complex solution.
A: A different way:
$$2^n\cos^n x=(e^{ix}+e^{-ix})^n=\sum_{k=0}^n\binom nke^{ikx}e^{-i(n-k)x}=\sum_{k=0}^n\binom nke^{i(2k-n)x}\\
=\sum_{k=(n+1)/2}^n\binom nk\cos((2k-n)x)$$
by grouping the terms of opposite exponents (for odd $n$; the case of even $n$ is very similar).
Then the antiderivative is
$$\sum_{k=(n+1)/2}^n\binom nk\frac{\sin((2k-n)x)}{2n-k}+C.$$
