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

1 Answer
Andrea S.
Jan 24, 2017

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

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