integration of sin^7 x cos^5 x

We'll solve this integral using substitution
technique.

Let cos x = t => - sin x dx =
dt

We'll use Pythagorean identity to write (sin x)^2, with
respect to (cos x)^2:

(sin x)^2 = 1 - (cos
x)^2

We'll raise to the cube both
sides:

(sin x)^6 = [1 - (cos
x)^2]^3

`int` (sin x)^6*(cos x)^5* sin xdx =`int` [1 - (cos
x)^2]^3*(cos x)^5* sin xdx

We'll re-write the integral
using the new variable t

`int` -
(1-t^2)^3*t^5*dt

We'll expand the
binomial:

`int` -(1 - 3t^2 + 3t^4 -
t^6)*t^5*dt

`int` (-t^5 + 3t^7 - 3t^9 + t^11)dt = -t^6/6 +
3t^8/8 - 3t^10/10 + t^12/12 + C

`int` (sin
x)^7*(cos x)^5 dx = -(cos x)^6/6 + 3(cos x)^8/8 - 3(cos x)^10/10 + (cos x)^12/12 +
C