How do you integrate int e^x sin ^2 x dx exsin2xdx using integration by parts?

1 Answer
Oct 10, 2017

The answer is =e^x/2-1/10e^xcos2x-1/5e^xsin2x+C=ex2110excos2x15exsin2x+C

Explanation:

We need

cos2x=1-2sin^2xcos2x=12sin2x, =>, sin^2x=(1-cos2x)/2sin2x=1cos2x2

Therefore,

I=inte^xsin^2xdx=int(e^x(1-cos2x)dx)/2I=exsin2xdx=ex(1cos2x)dx2

Now, perform the integration by parts

u=1-cos2xu=1cos2x, =>, u'=2sin2x

v'=e^x, =>, v=e^x

Therefore,

I=1/2(e^x(1-cos2x)-inte^x2sin2xdx)

=e^x/2(1-cos2x)-inte^xsin2xdx

Perform the integration by parts a second time

u=sin2x, =>, u'=2cos2x

v'=e^x, =>, v=e^x

I=1/2inte^xdx-1/2inte^xcos2x=e^x/2(1-cos2x)-e^xsin2x+2inte^xcos2x

Therefore,

5/2inte^xcos2x=e^x/2-e^x/2+1/2e^xcos2x+e^xsin2x

inte^xcos2x=2/5(e^x/2-e^x/2+1/2e^xcos2x+e^xsin2x)

=2/5(e^x)(1/2cos2x+sin2x)

inte^x(cos2x)=int(e^x)(1-2sin^2x)dx=e^x-2inte^xsin^2xdx

So,

e^x-2inte^xsin^2xdx=2/5(e^x)(1/2cos2x+sin2x)

inte^xsin^2xdx=e^x/2-1/10e^xcos2x-1/5e^xsin2x+C