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:

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

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