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I'm trying to find the integral of $e^{2\sin x}$. I tried using integration by parts and u-substitution but neither seems to work in the end. Would someone be able to take me through how to do this?

I was given this question by a classmate who found it but couldn't do it and we're in an extension maths class in high school. Going by the comments, i'm guessing this is beyond the scope of what we can do...

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    $\begingroup$ This integral is not elementary. The antiderivative of $\exp\bigl(2\sin x\bigr)$ cannot be written explicitly in elementary functions. $\endgroup$ – martini May 7 '15 at 11:22
  • $\begingroup$ What leads you to believe that a closed form for the integral exists? $\endgroup$ – Omnomnomnom May 7 '15 at 11:23
  • $\begingroup$ To find the definite integral, you can use contour integration. Or, the taylor expansion of e^2sinx, however, you can't really find an exact antiderivative among the usual suspects $\endgroup$ – user237393 May 7 '15 at 11:25
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As already said in comments, the antiderivative cannot be expressed in terms of elementary functions. So, may be, you could use a Taylor series, using $$e^y=1+y+\frac{y^2}{2}+\frac{y^3}{6}+O\left(y^4\right)$$ and replace $y$ by $2\sin(x)$. This will give $$e^{2\sin(x)}=1+2 \sin (x)+2 \sin ^2(x)+\frac{4 }{3}\sin ^3(x)+\cdots$$ and you are then left with a linear combination of integrals $$I_n=\int \sin^n(x)\, dx$$ for which exist at least a reduction formula $$I_n=-\frac 1n\sin^{n-1}(x)\cos(x)+\frac{n-1}n I_{n-2}$$

Edit

egreg made a good point in comments. When I wrote (in my initial answer) that the antiderivative does not exist, it was a bad shortcut to say that the antiderivative cannot be expressed in terms of elementary functions. Since I did not want to hide my mistake (not to say more), I did not modify the text. I change it now and let egreg the merit of pointing my bad wording.

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  • $\begingroup$ "Does not exist?" I can easily write one: $\int_0^x\exp(2\sin t)\,dt$. You probably are meaning something else. $\endgroup$ – egreg May 7 '15 at 11:42
  • $\begingroup$ @egreg. You are right ! I took a shortcut to mean that the antiderivative cannot be expressed in terms of elementary functions. I shall add it to my answer. Thanks for the precision. $\endgroup$ – Claude Leibovici May 7 '15 at 11:44
  • $\begingroup$ Why not changing the bad wording? Even textbooks are plagued with assertions such as “the antiderivative of $f(x)=e^{-x^2}$ does not exist”. Better not adding to the confusion, in my opinion. $\endgroup$ – egreg May 7 '15 at 11:48
  • $\begingroup$ I did not want to hide my stupidity and let you the credit of the good remark. I shall precise it. $\endgroup$ – Claude Leibovici May 7 '15 at 11:53
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Although the integrand does not possess a closed form anti-derivative in terms of elementary functions $($see Liouville's theorem and the Risch algorithm for more information$)$, its definite

counterpart is $\displaystyle\int_0^\tfrac\pi2e^{a\sin x}~dx~=~\int_0^\tfrac\pi2e^{a\cos x}~dx~=~\frac\pi2\Big[I_0(a)+L_0(a)\Big]$, where I and L are

the Bessel and Struve functions, respectively.

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  • $\begingroup$ For more information, see here. $\endgroup$ – Lucian May 7 '15 at 16:52

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