Note that: cos^3x = cosx xx cos^2x, then use the trigonometric identity:
cos^2x = (1-sin^2x)
Now:
int cos^3x dx = int cosx cos^2x dx = int cosx(1-sin^2x)dx
Using the linearity of the integral:
int cos^3x dx = int cosx dx - int sin^2x cosx dx
The first is a known integral:
int cosxdx = sinx +C
In the second substitute t = sinx so that dt = cosx dx and you have:
int sin^2x cosx dx = int t^2dt = t^3/3 +C = sin^3x/3+C
Finally:
int cos^3x dx = sinx-(sin^3x)/3+C
Second question:
int_0^pi sinx cos(cos(x)) dx
First solve the indefinite integral by substituting:
t= cosx
dt = -sinx dx
so that:
int sinx cosx(cosx)dx = - int costdt = -sint +C = -sin(cos(x)) + C
Now:
int_0^pi sinx cos(cos(x)) dx = [-sin(cos(x))]_0^pi
int_0^pi sinx cos(cos(x)) dx = sin(cos(0)) - sin(cos(pi))
int_0^pi sinx cos(cos(x)) dx = sin(1) - sin(-1)
but sinx is an odd function so sin(-1) = -sin(1) and then:
int_0^pi sinx cos(cos(x)) dx = 2sin(1)
