How would I integrate this:

$$-\int e^{\cos(t)}\sin(\sin(t)+t)\,dt $$

I have tried several methods but can't seem to work this out.

  • $\begingroup$ Which methods have you tried? $\endgroup$ – graydad Nov 20 '14 at 20:59
  • $\begingroup$ I original attempted it by inspection, then I tried integration by parts which got extremely messy. I tried substitution by I'm not very good at that $\endgroup$ – smith Nov 20 '14 at 21:00
  • $\begingroup$ I would play around with the identity $\sin(x) = \frac{1}{2}[e^{ix}-e^{-ix}]$. $\endgroup$ – graydad Nov 20 '14 at 21:01
  • $\begingroup$ I hadn't considered this thank you $\endgroup$ – smith Nov 20 '14 at 21:01
  • 1
    $\begingroup$ expand $\sin(\sin(t)+t)$ $\endgroup$ – Dr. Sonnhard Graubner Nov 20 '14 at 21:04

If you are allowed to use complex number, this integral can be integrated by repeat application of the Euler's formula $$e^{i\theta} = \cos(\theta) + i\sin\theta$$

Up to integration constant, the integral is equal to:

$$\begin{align}\int -e^{\cos t}\sin(\sin t + t) dt &= - \int e^{\cos t}\Im\left[e^{i(\sin t + t)}\right] dt \stackrel{\color{blue}{[1]}}{=} - \Im \left[\int e^{\cos t + i(\sin t + t)} dt \right]\\ &= - \Im \left[\int e^{e^{it}} e^{it} dt\right] = \Im \left[ i\int e^{e^{it}} d e^{it}\right]\\ &= \Re\left[ e^{e^{it}} \right] = \Re\left[ e^{\cos t + i\sin t} \right]\\ &= \, e^{\cos t} \cos(\sin t) \end{align} $$


  • $\color{blue}{[1]}$ - At this step, the integral is an ordinary integral of a complex valued function along the real axis. There is no issue of moving the $\Im[\cdots]$ outside the integral sign.
  • $\begingroup$ But $\operatorname{Im}\int f(z) \, \mathrm dz = \int \operatorname{Im} f(z) \, \mathrm dz$ is false in general? Consider $\int \operatorname{Log} z \, \mathrm dz$. $\endgroup$ – GFauxPas Nov 20 '14 at 22:30
  • $\begingroup$ @GFauxPax This is not complex contour integral, it is integral of a complex valued function along real axis. $\int \Im\left[f\right]dz = \int \Im\left[f dz\right] = \Im\left[\int f dz\right]$ does work in such cases. $\endgroup$ – achille hui Nov 20 '14 at 22:41
  • $\begingroup$ So you can always move $\operatorname{Im}$ and $\operatorname{Re}$ into and out of the integral if $z$ is wholly real? $\endgroup$ – GFauxPas Nov 20 '14 at 22:43
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    $\begingroup$ @GFauxPas Yes. you can move $\Im$ and $\Re$ into and out if $z$ is always real. $\endgroup$ – achille hui Nov 20 '14 at 22:51

Note that $$e^{\cos t} \sin(\sin(t) + t) = e^{\cos t} \left( \sin(\sin t)\cos t + \cos(\sin t)\sin t \right)$$ which is equal to $$-\left(e^{\cos t}(\cos t)'\cos(\sin t) + e^{\cos t}(\cos(\sin t))'\right)$$

Got it from here?

  • $\begingroup$ I don't understand the first line? $\endgroup$ – smith Nov 20 '14 at 21:13
  • $\begingroup$ The first line is applying this identity $$\sin(a + b) = \sin a \cos b + \cos a \sin b$$ where $a = \sin t, b = t$ $\endgroup$ – Simon S Nov 20 '14 at 21:14
  • $\begingroup$ Next one can use the formulae $$\int e^{g(x)}[f'(x) + g'(x)f(x)] dx = f(x) e^{g(x)}+C$$ which is proved here math.stackexchange.com/questions/1472748/… $\endgroup$ – Idris Feb 19 '17 at 14:44

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