How do you integrate #int x cos 3 x dx # using integration by parts?

1 Answer
Jan 10, 2016

Step by step working is shown below on how to go about integration by parts.

Explanation:

#intxcos(3x)dx#

Integration by parts rule is

#int udv = uv-int vdu#

First we select #u#
Using a mnemonic ILATE which stands for Inverse, Logarithmic, Algebraic, Trigonometric and Exponential we select #u# in the preferred order given by ILATE.

In our problem, we can select #u=x# this is because Algebraic function is preferred above the Trigonometric.

#u=x#
Differentiating with respect to #x#
#du = dx#

Once #u# is selected remaining terms for the #dv#

#dv=cos(3x)dx#

To find #v# we have to integrate this function.

#v=int dv = int cos(3x)dx#

#v=sin(3x)/3#

Using the integration part rule

#int udv = uv-int vdu#

#intxcos(3x)dx = 1/3xsin(3x) - int(sin(3x)/3)dx#

#intxcos(3x)dx = 1/3xsin(3x) +1/9cos(3x)+C#