How do you integrate #int x^2 cos x^2 dx # using integration by parts?

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Answer 1 · 2017-01-13T14:20:48

#int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C#

When we integrate by parts a function of the form:

#x^nf(x)#

we normally choose #x^n# as the integral part and #f(x)# as the differential part, so that in the resulting integral we have #x^(n-1)#

In this case however

#cos^2xdx#

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))/2#

#int x^2cos^2xdx = int x^2(1+cos(2x))/2dx = int x^2/2dx +int x^2/2 cos(2x)dx#

Now we can solve the first integral directly:

#int x^2/2dx = x^3/6#

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#

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 +C#

Putting it together:

#int x^2cos^2xdx = x^3/6 +(x^2sin(2x))/4 + (xcos(2x))/4 -1/8sin2x +C#