Integrate (e^x)(5sin(2x))?

Question

Don't answer if you don't know what I'm talking about!

Answer 1

As several have already mentioned, you can integrate by parts, but its really so much simpler. sin(2x) = (e^i2x - e^-i2x)/2i. The problem now simplifies to

5/(2i) Int.[exp((1+2i)x) - exp((1-2i)x)] =
5/(2i) [[ Int.[exp((1+2i)x)] - Int.[exp((1-2i)x)] ]]
Solving, we have

5/(2i) [[ 1/(1+2i) [exp((1+2i)x)] - (1-2i) [exp((1-2i)x)] ]] =
5/(2i-4) [exp((1+2i)x)] - 5/(2i+4) [exp((1-2i)x)] =
5(2i+4)/-20 [e^x * exp(2ix)] - 5(2i-4)/-20 [e^x * exp(-2ix)] =
-e^x*[[ (i+2)/2 [cos(2x)+isin(2x)] - (i-2)/2 [cos(-2x)+isin(-2x)] ]] =
-(e^x)/2*[[ (i+2) [cos(2x)+isin(2x)] - (i-2) [cos(2x)-isin(2x)] ]] =
-(e^x)/2*[[ ((i+2)-(i-2))cos(2x)+((i+2)--(i-2))isin(2x) ]] =
-(e^x)/2*[[ 4cos(2x)+(-2sin(2x) ]] =
-(e^x)*[[ 2cos(2x) - sin(2x) ]]

Note that while there may seem to be alot of algebra, the actual integration was exceedingly simple. Given the choice, I'll take hard algebra over hard calculus any day.
If you are feeling masochistic you could always use the series expansions....

Answer 2

Try letting e^x∙dx = dv and 5sin(2x) = u. The do integration by parts.

integral u∙dv = u∙v - integral v∙du


Doug

Answer 3

Mathematica will do this. They have an integrator online at http://www.wolfram.com

Having taken years of calculus, I see the above problem as a classic integration by parts. Rather than waste sheets of paper performing the integral, Mathematica will give the answer in spilt second.

Answer 4

do it by parts.
but you will have to do it at least twice, then you will get some sort of equation in the integral you are looking for.

Answer 5

Dude, buy a TI-92 Calculator and all you have to do is type that in and it solves it for you. It gives you the steps and everything.