How do you integrate #int e^x sin x ^2 dx # using integration by parts?

1 Answer
Apr 11, 2017

If this is what you meant, then:

#inte^xsin^2(x)dx=1/2e^x-1/10e^xcos(2x)-1/5e^xsin(2x)+C#

Explanation:

If you meant #inte^xsin^2(x)dx#, then this can be integrated.

We have to use the identity #sin^2(x)=1/2(1-cos(2x))#. The integral is then equal to:

#I=inte^xsin^2(x)dx=inte^x(1/2(1-cos(2x)))dx#

#color(white)I=1/2inte^xdx-1/2inte^xcos(2x)dx#

The first integral is elementary. We'll call the second one #J# and find it by itself:

#I=1/2e^x-1/2J#

For #J=inte^xcos(2x)dx#, we should use integration by parts. Let:

#{(u=cos(2x),=>,du=-2sin(2x)dx),(dv=e^xdx,=>,v=e^x):}#

So:

#J=e^xcos(2x)+int2e^xsin(2x)dx#

Perform integration by parts again.

#{(u=2sin(2x),=>,du=4cos(2x)dx),(dv=e^xdx,=>,v=e^x):}#

Then:

#J=e^xcos(2x)+2e^xsin(2x)-4inte^xcos(2x)dx#

Notice that #J# has reappeared on the right-hand side. We can add #4J# to both sides of the equation and then solve for #J#:

#5J=e^xcos(2x)+2e^xsin(2x)#

#J=1/5e^xcos(2x)+2/5e^xsin(2x)#

Returning to our original expression:

#I=1/2e^x-1/2J#

#color(white)I=1/2e^x-1/2(1/5e^xcos(2x)+2/5e^xsin(2x))#

#color(white)I=1/2e^x-1/10e^xcos(2x)-1/5e^xsin(2x)+C#