How do you integrate #(e^x)sin4x dx#?

Question

Emily Ammerman · Accepted Answer

This integral is cyclic and requires two iterations using the integration by parts theorem, which states:

#intf(x)g'(x)dx=f(x)g(x)-intg(x)f'(x)dx#

We can assume that #f(x)=sin4x# and #g'(x)dx=e^xdx#,

#f'(x)dx=cos4x*4dx#

#g(x)=e^x#,

so:

#I=e^xsin4x-inte^xcos4x*4dx=#

#=e^xsin4x-4inte^xcos4xdx=(1)#.

And now once more:

#f(x)=cos4x# and #g'(x)dx=e^xdx#,

#f'(x)dx=-sin4x*4dx#

#g(x)=e^x#.

So:

#(1)=e^xsin4x-4[e^xcos4x-inte^x(-sin4x*4)dx]=#

#=e^xsin4x-4e^xcos4x-16inte^xsin4xdx#.

At last, we are able to write:

#I=e^xsin4x-4e^xcos4x-16IrArr#

#17I=e^xsin4x-4e^xcos4xrArr#

#I=e^x/17(sin4x-4cos4x)+c#.

Ezra Adams · Answer

undefined To integrate $ \int e^x \sin(4x) \, dx $, you can use integration by parts. Let $ u = e^x $ and $ dv = \sin(4x) \, dx $. Then, $ du = e^x \, dx $ and $ v = -\frac{1}{4} \cos(4x) $. Applying the integration by parts formula, you get:

$$ \int e^x \sin(4x) \, dx = -\frac{1}{4} e^x \cos(4x) - \frac{1}{4} \int e^x \cos(4x) \, dx $$

Now, integrate $ \int e^x \cos(4x) \, dx $ using integration by parts again. Let $ u = e^x $ and $ dv = \cos(4x) \, dx $. Then, $ du = e^x \, dx $ and $ v = \frac{1}{4} \sin(4x) $. Substituting these values, you get:

$$ \int e^x \cos(4x) \, dx = \frac{1}{4} e^x \sin(4x) - \frac{1}{4} \int e^x \sin(4x) \, dx $$

Now, substitute this result back into the original integral equation and solve for $ \int e^x \sin(4x) \, dx $. You'll get:

$$ \int e^x \sin(4x) \, dx = -\frac{1}{4} e^x \cos(4x) - \frac{1}{16} e^x \sin(4x) + C $$

where $ C $ is the constant of integration.