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

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Answer 1 · 2017-04-11T21:38:24

If this is what you meant, then:

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

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:

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

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

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

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