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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!

amWhy
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NeilRoy
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    The graphics of the integrand function is symmetric with respect to the point $(\pi,0)$, hence you are right, the integral is zero. – Jack D'Aurizio Aug 29 '14 at 15:36

3 Answers3

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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

lab bhattacharjee
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Here's another way of looking at it:

$$\begin{align} \int_{0}^{2\pi}e^{\cos{x}}\cos{x}\sin{x}\,\mathrm{d}x &=\int_{-\pi}^{\pi}e^{\cos{x}}\cos{x}\sin{x}\,\mathrm{d}x\\ &=\int_{-\pi}^{\pi}e^{\cos{x}}\frac{\sin{(2x)}}{2}\,\mathrm{d}x \end{align}$$

David H
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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 $$\int \mathrm{e}^{\cos x}\sin x~\cos x~\mathrm{dx} = \int - \ln u~\mathrm{d}u = u(1-\ln u) + C = \mathrm{e}^{\cos x}(1-\cos x)+C$$

Fly by Night
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