Integration by parts

Let I(n) = Int (cos x)^n dx

Int = integral symbol (looks like a snake)

Let du/dx = cos x, v = (cos x)^(n-1)
then u = sin x, dv/dx = -(n-1)(cos x)^(n-2) sin x

I(n) = Int (cos x)^(n-1) cos x dx
= (cos x)^(n-1) sin x - Int -(n-1) (sin x)^2 (cos x)^(n-2) dx (using formula of int. by parts)
= (cos x)^(n-1) sin x + (n-1) Int (sin x)^2 (cos x)^(n-2) dx
= (cos x)^(n-1) sin x + (n-1) Int [1 - (cos x)^2] (cos x)^(n-2) dx (you know the identity)
= (cos x)^(n-1) sin x + (n-1) Int (cos x)^(n-2) dx - (n-1) Int (cos x)^2 (cos x)^(n-2) dx
= (cos x)^(n-1) sin x + (n-1) Int (cos x)^(n-2) dx - (n-1) Int (cos x)^n dx
= (cos x)^(n-1) sin x + (n-1) I(n-2) - (n-1) I(n)

So,
I(n) = (cos x)^(n-1) sin x + (n-1) I(n-2) - (n-1) I(n)
I(n) + (n-1) I(n) = (cos x)^(n-1) sin x + (n-1) I(n-2)
n I(n) = (cos x)^(n-1) sin x + (n-1) I(n-2)
I(n) = [1/n] (cos x)^(n-1) sin x + [(n-1)/n] I(n-2)

I(0) = Int dx = x + c
I(1) = Int cos x dx = sin x + c