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)#
