How do you integrate #int e^(3x)cos(2x)# by integration by parts method?

Question

Paisley Johnson · Accepted Answer

I got: #9/13[e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)]+c#
BUT check my maths.

Ok let us try: #inte^(3x)cos(2x)dx=e^(3x)/3cos(2x)-int(e^(3x)/3)(-2sin(2x))dx=e^(3x)/3cos(2x)+2/3int(e^(3x))sin(2x)dx=# again: #=e^(3x)/3cos(2x)+2/3[e^(3x)/3sin(2x)-int(e^(3x)/3)(2cos(2x))dx]# So basically we get: #inte^(3x)cos(2x)dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)-4/9inte^(3x)(2cos(2x))dx# Now a trick.... Let us take #-4/9inte^(3x)(2cos(2x))dx# to the left of the #=# sign: #inte^(3x)cos(2x)dx+4/9inte^(3x)(2cos(2x))dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)# add the two integral on the left: #13/9inte^(3x)(2cos(2x))dx=e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)# and: #inte^(3x)(2cos(2x))dx=9/13[e^(3x)/3cos(2x)+2/9e^(3x)sin(2x)]+c#

Elijah Abram · Answer

undefined To integrate $ \int e^{3x}\cos(2x) $ using the integration by parts method, you would assign one function as $ u $ and the other as $ dv $. Let $ u = e^{3x} $ and $ dv = \cos(2x)dx $. Then, you would differentiate $ u $ and integrate $ dv $ to find $ du $ and $ v $, respectively. After that, you would apply the integration by parts formula:

$$ \int u \, dv = uv - \int v \, du $$

Substitute the values of $ u $, $ v $, $ du $, and $ dv $ into the formula, then evaluate the integral.