High school integral problem?

Question

i don't know how to write an integral function here but it's an integral in x. or in other words, an integral with dx in the end.

the question is=

(integral) e^x . sinx . sin3x dx

thanks.

it's really harddd and i've been struggling for minutes!

Answer 1

After quite a bit of writing (1 printer paper page, front and back) I came up with an answer. I am not going to type it all out, but if you want the work, email me and I will see if I can scan it in and send it to you (kind of messy hehe). Anyways, heres the answer I came up with:

(e^x Sin(x) Sin(3x) - 27/11 e^x Sin(x) Cos(3x) - (7/11) e^x Sin(3x) Cos(x) + (6/11) e^x Cos(x) Cos(3x)) / (-25) + C

I found it by using integration by parts multiple times until I was left with Integral of e^x Cos(x) Cos(3x) dx, and Integral of e^x Sin(x) Sin(3x) dx, I then used integration by parts to find the Integral of e^x Cos(x) Cos(3x) dx in terms of Integral of e^x Sin(x) Sin(3x) dx, and substituted that value back into where I was at with the original integration by parts.

I used Mathematica to check my answer (had my solution as a function, then found the solution with Mathematica and had that in another function). It says that they are not the same. Here is the solution that Mathematica gives:

1/170 e^x (17 Cos[2 x]-5 Cos[4 x]+34 Sin[2 x]-20 Sin[4 x])

Or, in another form, expanded out:

1/10 e^x Cos[x]^2-1/34 e^x Cos[x]^4+2/5 e^x Cos[x] Sin[x]-8/17 e^x Cos[x]^3 Sin[x]-1/10 e^x Sin[x]^2+3/17 e^x Cos[x]^2 Sin[x]^2+8/17 e^x Cos[x] Sin[x]^3-1/34 e^x Sin[x]^4

Then with terms collected:

-(1/170) e^x (-17 Cos[x]^2+5 Cos[x]^4-68 Cos[x] Sin[x]+80 Cos[x]^3 Sin[x]+17 Sin[x]^2-30 Cos[x]^2 Sin[x]^2-80 Cos[x] Sin[x]^3+5 Sin[x]^4)

I am not quite sure where I went wrong with my solution, and I have quite a bit of work to look through to try and figure it out, but the method I went through (and described above) is how you would go about finding the answer if you were to do it yourself.

Answer 2

that integral keeps repeating itself over and over and over again