When we integrate by parts a function of the form:
x^nf(x)xnf(x)
we normally choose x^nxn as the integral part and f(x)f(x) as the differential part, so that in the resulting integral we have x^(n-1)xn−1
In this case however
cos^2xdxcos2xdx
is not the differential of an «easy» function, so we first reduce the degree of the trigonometric function using the identity:
cos^2x = (1+cos(2x))/2cos2x=1+cos(2x)2
int x^2cos^2xdx = int x^2(1+cos(2x))/2dx = int x^2/2dx +int x^2/2 cos(2x)dx∫x2cos2xdx=∫x21+cos(2x)2dx=∫x22dx+∫x22cos(2x)dx
Now we can solve the first integral directly:
int x^2/2dx = x^3/6∫x22dx=x36
and the second by parts:
int x^2/2 cos(2x)dx = 1/4 int x^2 d(sin2x) = (x^2sin(2x))/4 -1/2 int xsin(2x)dx∫x22cos(2x)dx=14∫x2d(sin2x)=x2sin(2x)4−12∫xsin(2x)dx
and again:
1/2 int xsin(2x)dx = -1/4 int xd(cos2x) = -(xcos(2x))/4 + 1/4 int cos2xdx = -(xcos(2x))/4 +1/8sin2x +C12∫xsin(2x)dx=−14∫xd(cos2x)=−xcos(2x)4+14∫cos2xdx=−xcos(2x)4+18sin2x+C
Putting it together:
int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C∫x2cos2xdx=x36+x2sin(2x)4+xcos(2x)4−18sin2x+C
