# Thread: Problem with a complicated Integral

1. ## Problem with a complicated Integral

Hello,

I want to integrate $\int_{0}^{2\pi} \cos(t+a)\ \cos(b \cos(t)) \ e^{-c |\cos(t)|}\ dt$ where $a,b,c$ are constants. I tried in Matlab and Mathematica but both are unable to solve the integral. Is there a trick to simplify this integral to make it integrable? Any help is highly appreciated.

Thanks and Regards,
Ameya

2. Originally Posted by amrasa81
Hello,

I want to integrate $\int_{0}^{2\pi} \cos(t+a)\ \cos(b \cos(t)) \ e^{-c |\cos(t)|}\ dt$ where $a,b,c$ are constants. I tried in Matlab and Mathematica but both are unable to solve the integral. Is there a trick to simplify this integral to make it integrable? Any help is highly appreciated.

Thanks and Regards,
Ameya
cos(t+ a)= cos(t)cos(a)- sin(t)sin(a) so you can write this as
$cos(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}cos(t)dt- sin(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}sin(t)dt$

You can simplify the second integral, with the substitution u= cos(t), to $sin(a)\int_0^0 cos(bu)e^{-c|u|}du= 0$ simply because the limits of integration are the same. The first integral is a little more complicated but $\int_0^0 cos(u)e^{-|u|}\frac{u}{\sqrt{1- u^2}}du$ is still 0.

3. Originally Posted by HallsofIvy
cos(t+ a)= cos(t)cos(a)- sin(t)sin(a) so you can write this as
$cos(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}cos(t)dt- sin(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}sin(t)dt$

You can simplify the second integral, with the substitution u= cos(t), to $sin(a)\int_0^0 cos(bu)e^{-c|u|}du= 0$ simply because the limits of integration are the same. The first integral is a little more complicated but $\int_0^0 cos(u)e^{-|u|}\frac{u}{\sqrt{1- u^2}}du$ is still 0.
Hello HallsofIvy,

Thank you for your reply. So what essentially you are saying is that the value of my original integral is zero?

Regards,
Ameya

4. Originally Posted by HallsofIvy
cos(t+ a)= cos(t)cos(a)- sin(t)sin(a) so you can write this as
$cos(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}cos(t)dt- sin(a)\int_0^{2\pi} cos(bcos(t))e^{-c|cos(t)|}sin(t)dt$

You can simplify the second integral, with the substitution u= cos(t), to $sin(a)\int_0^0 cos(bu)e^{-c|u|}du= 0$ simply because the limits of integration are the same. The first integral is a little more complicated but $\int_0^0 cos(u)e^{-|u|}\frac{u}{\sqrt{1- u^2}}du$ is still 0.
Ok, I see where the problem is with the substitution. I think your substitution is invalid because within the interval $(0,2\pi), x = \cos t$ does not satisfy the bijection, and hence, I have to break this integral up. Anyway, thanks for your effort again.