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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