Question

How do you integrate `int x^2cos3x` by integration by parts method?

Answer 1

`int x^2cos3x dx = (x^2sin3x)/3 +(2xcos3x)/9-(2sin3x)/27+C`

Explanation:

When integrating by parts a function in the form `f(x) = x^kg(x)` we normally take `x^k` as finite factor, so that at the next iteration the grade of `x` is lower.

So, we can note that `cos3x dx = 1/3d (sin3x)` and integrate by parts in this way:

`int x^2cos3x dx = 1/3 int x^2 d(sin3x) = (x^2sin3x)/3 - 2/3 int xsin3x(dx)`

The resulting integral can also be resolved by parts:

`int xsin3x(dx) = -1/3 int xd(cos3x) = -(xcos3x)/3 + 1/3 int cos3xdx = -(xcos3x)/3 + (sin3x)/9 + C`

Putting it all together:

`int x^2cos3x dx = (x^2sin3x)/3 +(2xcos3x)/9-(2sin3x)/27+C`