Integration of $\int_0^{2\pi}(e^{\cos x}\cos x\sin x) \,dx$ Can anyone please help me with this integration:
$$\int_0^{2\pi}(e^{\cos x}\cos x\sin x)\,dx$$
I am getting the answer as $0$ by using simply the properties of definite integral
BUT the answer as told by my teacher is $2\pi$
Please help. Thanks!
 A: Using $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$
If $\displaystyle  f(x)=e^{\cos x}\sin x\cos x,$
$\displaystyle f(2\pi+0-x)=e^{\cos(2\pi+0-x)}\sin(2\pi+0-x)\cos(2\pi+0-x)=e^{\cos x}\cos x(-\sin x)=-f(x)$

Also putting $\cos x=u,$
$$\int_0^{2\pi}e^{\cos x}\sin x\cos x=-\int_1^1e^u\cdot u\ du$$
The indefinite part can be managed using integration by part.
But as both the upper & the lower bounds are same, the result has to be zero
A: Here's another way of looking at it:
$$\begin{align}
\int_{0}^{2\pi}e^{\cos{x}}\cos{x}\sin{x}\,\mathrm{d}x
&=\int_{-\pi}^{\pi}e^{\cos{x}}\cos{x}\sin{x}\,\mathrm{d}x\\
&=\int_{-\pi}^{\pi}e^{\cos{x}}\frac{\sin{(2x)}}{2}\,\mathrm{d}x
\end{align}$$
A: If you make the simple substitution $u = \mathrm{e}^{\cos x}$ then you'll find that $\mathrm{d}u = -\mathrm{e}^{\cos x}\sin x~\mathrm{d}x$. Moreover, $\cos x = \ln u$. When $x=0$, $u=\mathrm{e}^{\cos 0} = \mathrm{e}$ and when $x=2\pi$, $u=\mathrm{e}^{\cos 2\pi} = \mathrm{e}$. Hence
$$\int_0^{2\pi} \mathrm{e}^{\cos x}\sin x~\cos x ~ \mathrm{d}x = \int_0^0 - \ln u~\mathrm{d}u=0$$
You don't need to evaluate the integral because the upper and lower limits are the same, so the integral will evaluate to zero. If you wanted to evaluate the integral you'd see that
$$\int \mathrm{e}^{\cos x}\sin x~\cos x~\mathrm{dx} = \int - \ln u~\mathrm{d}u = u(1-\ln u) + C = \mathrm{e}^{\cos x}(1-\cos x)+C$$
