How do you integrate #sin(x)*(e)^(2 x) dx#?

1 Answer
Jul 1, 2016

Answer:

Use integration by parts (IBP).

Solution: # int e^(2x)sin(x)dx = 2/5e^(2x)sin(x) - 1/5e^(2x)cos(x) + C#

Explanation:

To obtain the formula for IBP, we start with the product rule:

#(u*v)' = u'*v + u*v'# where u and v are functions of x in this case

Rearranging:

#u*v' = (u*v)' - u'*v#

Integrating:

#int u*v' dx = u*v - int u'*v dx#

For this integral, we are going to set #u(x) = sin(x)# and #v'(x) = e^(2x)#

#u'(x) = cos(x)# and #v(x) = 1/2e^(2x)#

So, #int e^(2x)sinxdx = 1/2e^(2x)sin(x) - 1/2*int cos(x)e^(2x) dx#

Now do IBP again on the second term:

#u(x) = cos(x)# and #v'(x) = e^(2x)#
#u'(x) = -sin(x)# and #v(x) = 1/2e^(2x)#

#int e^(2x)sin(x)dx = 1/2e^(2x)sin(x) - 1/4e^(2x)cos(x) - 1/4*int e^(2x)sin(x)dx#

As you can see, the integral on the LHS is the same as the integral on the RHS so we collect like terms and obtain:

#5/4*int e^(2x)sin(x)dx = 1/2e^(2x)sin(x) - 1/4e^(2x)cos(x)#

# int e^(2x)sin(x)dx = 2/5e^(2x)sin(x) - 1/5e^(2x)cos(x) + C#