I am trying to prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$
This problem is a classic, but I seem to be missing one step or the understanding of two steps which I will outline below.
$I_n := \int\cos^n x \ dx = \int\cos^{n-1} x \cos x \ dx \tag{1}$
First question: why rewrite the original instead of immediately integrating by parts of $\int \cos^n x \ dx$?
Integrate by parts with $u = \cos^{n-1} x, dv = \cos x \ dx \implies du = (n-1)\cos^{n-2} x \cdot -\sin x, v = \sin x$
which leads to
$I_n = \sin x \ \cos^{n-1} x +\int\sin^2 x (n-1) \ \cos^{n-2} x \ dx \tag{2}$
Since $(n-1)$ is a constant, we can throw it out front of the integral:
$I_n = \sin x \ \cos^{n-1} x +(n-1)\int\sin^2 x \ \cos^{n-2} x \ dx\tag{3}$
I can transform the integral a bit because $\sin^2 x + cos^2 x = 1 \implies \sin^2 x = 1-\cos^2 x$
$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$
According to Wikipedia as noted here, this simplifies to:
$I_n = \sin x \ \cos^{n-1} x + (n-1) \int \cos^{n-2} x \ dx - (n-1)\int(\cos^n x) \ dx \tag{5}$
Question 2: How did they simplify the integral of $\int(1-\cos^2 x) \ dx$ to $\int(\cos^n x) \ dx$?
Assuming knowledge of equation 5, I see how to rewrite it as
$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x - (n-1) I_{n} \tag{6}$
and solve for $I_n$. I had tried exploiting the fact that $\cos^2 x = \frac{1}{2} \cos(2x) + \frac{1}{2} $
and trying to deal with $\int 1 \ dx - \int \frac{1}{2} \cos (2x) + \frac{1}{2} \ dx$
which left me with $\frac{x}{2} - \frac{1}{4} \sin(2x)$ after integrating those pieces. Putting it all together I have:
$I_n = \sin x \ \cos^{n-1} x + (n-1) I_{n-2} x \left(-(n-1) (\frac{x}{2} - \frac{1}{4} \sin 2x) \right) \tag{7}$
but I'm unsure how to write the last few terms as an expression of $I_{something}$ to get it to match the usual reduction formula of
$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$