# Definite integral of $\int e^{-\sin^2x}\cos\Bigl(6x-\frac{\sin(2x)}{2}\Bigr)dx$

I have the integral with $$\sin()$$ sum expression in $$\cos()$$ argument: $$\int_0^{2\pi}e^{-\sin^2x}\cos\Bigl(6x-\frac{\sin(2x)}{2}\Bigr)dx.$$ Can anyone please explain an algorithm for solving it?

I've tried so far: Weierstrass substitution (if I can use it with $$\int \cos(f(x))dx$$), Euler formula, integration by parts, formula of $$\cos(x-y)$$ and $$\sin(2x)$$ formula and everything not seems to work or I make actions not in the right order. Need some fresh view on the problem.

Thanks!

• The function is $\pi$ periodic, which means you can integrate on any interval of length $\pi$ no matter what start and end points are. Jul 16, 2021 at 19:23
• That's interesting remark, but how... will it help? Do you suggest any specific interval except 0 to $\pi$ or $\pi$ to $2\pi$? Jul 16, 2021 at 19:41
• Do you need to know the "exact" value of the integral? Or will a series representation be enough? Or do you want to find an approximate value of it?
– MasB
Jul 16, 2021 at 19:58
• First of all I've thought about simplifying it somehow. I don't need super-precisive value as answer, approximate representation would be enough in form of sum of fractures may be, but I need to know how to transform original to get it. Jul 16, 2021 at 20:06

$$\begin{split} \int_0^{2\pi}e^{-\sin^2x}\cos\Bigl(6x-\frac{\sin(2x)}{2}\Bigr)dx &= \Re \int_0^{2\pi}e^{-\sin^2x}e^{6ix}e^{-i\frac{\sin(2x)}2}dx \\ &= \Re \int_0^{2\pi}e^{-\frac 1 2}e^{\frac{\cos (2x)}2}e^{6ix}e^{-i\frac{\sin(2x)}2}dx \\ &= e^{-\frac 1 2}\Re \int_0^{2\pi}e^{6ix}e^{\frac{e^{-2ix}}2}dx \\ &= e^{-\frac 1 2}\Re \int_0^{2\pi}e^{6ix}\left (\sum_{n\geq 0} \frac 1 {n!}\left ( \frac{e^{-2ix}} 2\right)^n \right)dx \\ &=e^{-\frac 1 2}\Re \sum_{n\geq 0} \frac 1 {n!}\frac {1}{2^n} \int_0^{2\pi}e^{6ix}e^{-2inx}dx\\ &=e^{-\frac 1 2} \left ( \frac 1 {3!} \frac {1}{2^3} 2\pi\right )\\ &=\frac{e^{-\frac 1 2}\pi}{24} \end{split}$$ (In case you're wondering how the sum disappeared, notice that all terms in the sum are equal to 0 except for when $$n=3$$).
• I'm sorry, but could you explain please how did you get $e^{6ix}e^{-isin(2x)/2}$ in more details? Thats like $cos(f(x))=e^{i(f(x))}+e^{-i(f(x))}/2$? Jul 16, 2021 at 20:35
• Using $\cos (a+b) = \Re e^{i(a+b)}=\Re (e^{ia} e^{ib})$ where $\Re$ is the real part. Jul 16, 2021 at 20:37
• quora.com/… Oh yeah, I've missed that $\cos(x)=\cos(A+iB)$ You are seem right Jul 16, 2021 at 21:56
• I wonder if it is true that: for all $\alpha,\beta\in\mathbb Z$ we have the following. $$I(\alpha,\beta)=\int_0^{\alpha\pi}e^{-\sin^2 x}\cos(2\beta x - \sin x\cos x)\,\mathrm dx$$ $$=\frac{\pi}{\sqrt e}\cdot\frac{\alpha}{2^\beta \beta !}$$ where of course $\beta < 0$ yields $\infty$. If true, we would have $$I(\alpha,\beta)=\frac 1{2\beta}I(\alpha,\beta - 1)$$ $$=\frac 1{2^n\beta(\beta-1)\cdots (\beta - n + 1)}I(\alpha,\beta - n)$$ Jul 19, 2021 at 4:35
• The same calculation I used can be employed to prove that, if $\beta\in\mathbb N$, $$\int_0^{2\pi}e^{-\sin^2 x}\cos(2\beta x - \sin x \cos x)dx = \frac{\pi}{\sqrt{e}}\frac 1 {\beta!}\frac 1 { 2^{\beta -1}}$$ I'm not sure about the general formula you're proposing. Jul 19, 2021 at 6:44