On $\int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=2\pi$ Here's my attempt at an integral I found on this site.
$$\int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=2\pi$$
I'm not asking for a proof, I just want to know where I messed up
Recall that, for all $x$,
$$e^x=\sum_{n\geq0}\frac{x^n}{n!}$$
And  $$\cos x=\sum_{n\geq0}(-1)^n\frac{x^{2n}}{(2n)!}$$
Hence we have that 
$$
\begin{align}
\int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=&\int_0^{2\pi}\bigg(\sum_{n\geq0}\frac{\cos^n2x}{n!}\bigg)\bigg(\sum_{m\geq0}(-1)^m\frac{\sin^{2m}2x}{(2m)!}\bigg)\mathrm{d}x\\
=&\sum_{n,m\geq0}\frac{(-1)^m}{n!(2m)!}\int_0^{2\pi}\cos(2x)^n\sin(2x)^{2m}\mathrm{d}x\\
=&\frac12\sum_{n,m\geq0}\frac{(-1)^m}{n!(2m)!}\int_0^{4\pi}\cos(t)^n\sin(t)^{2m}\mathrm{d}t\\
\end{align}
$$
The final integral is related to the incomplete beta function, defined as 
$$B(x;a,b)=\int_0^x u^{a-1}(1-u)^{b-1}\mathrm{d}u$$
If we define 
$$I(x;a,b)=\int_0^x\sin(t)^a\cos(t)^b\mathrm{d}t$$
We can make the substitution $\sin^2t=u$, which gives 
$$
\begin{align}
I(x;a,b)=&\frac12\int_0^{\sin^2x}u^{a/2}(1-u)^{b/2}u^{-1/2}(1-u)^{-1/2}\mathrm{d}u\\
=&\frac12\int_0^{\sin^2x}u^{\frac{a-1}2}(1-u)^{\frac{b-1}2}\mathrm{d}u\\
=&\frac12\int_0^{\sin^2x}u^{\frac{a+1}2-1}(1-u)^{\frac{b+1}2-1}\mathrm{d}u\\
=&\frac12B\bigg(\sin^2x;\frac{a+1}2,\frac{b+1}2\bigg)\\
\end{align}
$$
Hence we have a form of our final integral:
$$
\begin{align}
I(4\pi;2m,n)=&\frac12B\bigg(\sin^24\pi;\frac{2m+1}2,\frac{n+1}2\bigg)\\
=&\frac12B\bigg(0;\frac{2m+1}2,\frac{n+1}2\bigg)\\
=&\frac12\int_0^0t^{\frac{2m-1}2}(1-t)^{\frac{n-1}2}\mathrm{d}t\\
=&\,0
\end{align}
$$
Which implies that 
$$\int_0^{2\pi}e^{\cos2x}\cos(\sin2x)\ \mathrm{d}x=0$$
Which is totally wrong. But as far as I can tell, I haven't broken any rules. Where's my error, and how do I fix it? Thanks.
 A: You cannot substitute $u=\sin^2t$. As $t$ ranges from $0$ to $2\pi$, this is not a one-to-one relationship.
It's like if you subbed $u=x^2$ in $$\int_{-1}^1x^2\,dx$$ You would get an integral from $u=1$ to $u=1$, which would be $0$ even though the integral is clearly nonzero.
A: To solve the integral, you may consider $$\int_{C} \frac{e^z}{z}dz$$
where $C$ is a unit circle, and see its real part.
A: If you want a full solution of the integral using complex analysis, going along the lines of what Seewoo Lee recommend, you can solve the integral as follows:
First consider the integral $$\int_{C} \frac{e^z}{z}dz$$
where $C$ is the unit circle oriented counter-clockwise in the complex plane. Using the Cauchy Integral formula (from complex analysis) you can find that $$\int_{C} \frac{e^z}{z}dz = 2\pi i \phantom{------}(1)$$
Now we will directly integrate the above integral by parametrising $C$. Let $z(t)=e^{2it}$ with $0 \le t \le \pi$ be the parametrisation of $C$. Then the integral works out as:
\begin{align}
\int_{C} \frac{e^z}{z}dz &= \int_0^{\pi} \frac{e^{e^{2it}}}{e^{2it}} 2ie^{2it} dt \\
&= 2i \int_0^{\pi} {e^{\cos(2t)+i\sin(2t)}}dt \\
&= 2i \int_0^{\pi} {e^{\cos(2t)}e^{i\sin(2t)}}dt \\
&= 2i \int_0^{\pi} {e^{\cos(2t)}(\cos(\sin(2t))+i\sin(\sin(2t))} dt \\
&= 2i \int_0^{\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt - 2 \int_0^{\pi} {e^{\cos(2t)}}(sin(sin(2t)) dt \phantom{-------} (2) \\
\end{align}
Equating imaginary parts of (1) and (2) we see that:
$$\int_0^{\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt = \pi$$
If we parametrise the curve initially with $\pi \le t \le 2\pi$ we would have ended up with:
$$\int_{\pi}^{2\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt = \pi$$
Thus,
\begin{align}
\int_{0}^{2\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt &= \int_{0}^{\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt + \int_{\pi}^{2\pi} {e^{\cos(2t)}}(\cos(\sin(2t))dt \\
&= \pi + \pi = 2\pi
\end{align}
A: you could try this:
$$I=\int_0^{2\pi}e^{\cos(2x)}\cos\left[\sin(2x)\right]dx$$
$$=\Re\left(\int_0^{2\pi}e^{\cos(2x)}\cos\left[\sin(2x)\right]dx+i\int_0^{2\pi}e^{\cos(2x)}\sin\left[\sin(2x)\right]dx\right)$$
$$=\Re\left(\int_0^{2\pi}e^{\cos(2x)}e^{i\sin(2x)}dx\right)$$
$$=\Re\left(\int_0^{2\pi}e^{\cos(2x)+i\sin(2x)}dx\right)$$
$$=\Re\left(\int_0^{2\pi}e^{e^{2ix}}dx\right)$$
and since:
$$e^y=\sum_{n=0}^\infty\frac{x^n}{n!}$$
we can say that:
$$e^{e^{2ix}}=\sum_{n=0}^\infty\frac{e^{2nix}}{n!}$$
and so:
$$I=\Re\int_0^{2\pi}\sum_{n=0}^\infty\frac{e^{2nix}}{n!}dx$$
$$=\Re\sum_{n=0}^\infty\left[\frac{e^{2nix}}{2ni.n!}\right]_0^{2\pi}$$
$$=\Re\sum_{n=0}^\infty\frac{e^{4\pi ni}-1}{2ni.n!}$$
but note that for all integers n, $$e^{4\pi ni}=1$$
so this summation may be hard to calculate (or wrong). This is probably due to the fact that the integral is between $0$ and $2\pi$, so the integral may need to be split up into several parts before it can be evaulated.
A: If by $\cos \sin (2 x)$ you really mean $\cos (2 x ) \sin (2 x)$, then the full function looks like

and the full integral is $0$.
If instead the integral is:
$$\int\limits_{x=0}^{2 \pi} e^{\cos (2 x)} \cos \left( \sin ( 2 x)\right)\ dx$$
the graph is:

and Mathematica gives the answer as $2 \pi$.
