Prove $\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$ 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 
$$\int\cos^n x \ dx = \frac{1}n \cos^{n-1}x \sin x + \frac{n-1}{n}\int\cos^{n-2} x \ dx$$
 A: I guess your problem is step four:
$$I_n = \sin x \ \cos^{n-1} x + (n-1)\int(1-\cos^2 x) \ \cos^{n-2} x \ dx \tag{4}$$
Note that 
$$\begin{align}\int(1-\cos^2 x) \ \cos^{n-2} x \ dx & = \int \cos^{n-2} x \ dx-\int \cos^2 x\; \cos^{n-2} x \ dx \\ & =\int \cos^{n-2} x \ dx-\int \cos^{n} x \ dx \end{align}$$
so we get
$$\begin{align}I_n & = \sin x \ \cos^{n-1} x + (n-1)\int \cos^{n-2} x \ dx-(n-1)\int \cos^{n} x \ dx \\ I_n & = \sin x \ \cos^{n-1} x + (n-1)I_{n-2}-(n-1)I_n \end{align}$$
or
$$nI_n = \sin x \ \cos^{n-1} x + (n-1)I_{n-2}$$
$$I_n =\frac{ \sin x \ \cos^{n-1} x}n + \frac{n-1}nI_{n-2}$$
A: How about verifying using differentiation? We are asked to show 
$$\int_0^x \cos^n t dt = \frac{1}{n} \cos^{n-1} x \sin x + \frac{n-1}{n} \int^x_0 \cos^{n-2} t dt + C$$
for some constant $C$. This is true iff the derivatives of each side are equal.
Differentiating the right side, we get using the product rule
$$RHS = \frac{n-1}{n} \cos^{n-2} x (-\sin x) \sin x + \frac{1}{n} \cos^{n-1} x \cos x + \frac{n-1}{n} \cos^{n-2} x,$$
then combining,
$$RHS = \frac{n-1}{n} (1-\sin^2 x) \cos^{n-2} x + \frac{1}{n} \cos^n x$$
and finally using $\cos^2 x + \sin^2 x = 1$,
$$RHS = \frac{n-1}{n}\cos^n x+\frac{1}{n} \cos^nx = \cos^n x,$$
which is the what we want.
