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When I asked this question in others forums, I took this answer.Is this true and how we can evaluate this given integral?

Let's assume, $$\int y dy=\int e^{\sin x} dx$$ And we have equality like that.In this case ,It'll be $$\frac{y^2}{2}=\int e^{\sin x} dx$$ Last equality means to $$\int \ln y \:dy=\int \sin x dx$$ Then we definitely know that $$\int \ln y\: dy=y(lny-1)$$ $$\Rightarrow$$ $$y(\ln y-1)=-\cos x+c$$ Then we get this equality; $$y=\frac{-\cos x+c}{\ln y-1}$$ But we want to find $y^2/2$;
$$\frac{y^2}{2}=\dfrac{\left(\frac{-\cos x+c}{\ln y-1}\right)^2}{2}=\int e^{\sin x} dx.$$

So we got the answer and this answer is a non-linear solution;

Micheal Brain Hurts
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    Wait a minute. Did you just take the logarithm of an integral by taking the logarithm of the integrand? I can't make any sense out of this. – Harald Hanche-Olsen Dec 04 '16 at 22:16
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    Why do you think there is a closed form for $\int e^{\sin x}\ dx$? –  Dec 04 '16 at 22:20
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    @Jack I am not sure, whether the OP assumes this. The integral containing the $y$ could have something to do with the required integral. But I admit that I cannot follow the steps – Peter Dec 04 '16 at 22:24
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    Replacing $e^{\sin x}$ with $e^{x^2}$ and following the same "method", can you see what you get? –  Dec 04 '16 at 22:30
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    There are several serious mistakes in your post. Firstly, you seem to be solving the equation $y \frac{dy}{dy} = e^{\sin x}$ without realizing it. Secondly, $\frac{y^2}{2}=\int e^{\sin x} dx$ does not imply $\int \ln y :dy=\int \sin x dx$. – Hans Engler Dec 04 '16 at 23:19
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    If you take the derivative of your answer, you won't get $e^{\sin x}$... This should tell you something.... – user209663 Dec 05 '16 at 02:30
  • Okey, thanx a lot of your interesting but still I don't find anything about solution or eveluation of this integral. Can you show/tell/guide some further information(s)? – Micheal Brain Hurts Dec 05 '16 at 13:24
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    Possible duplicate of https://math.stackexchange.com/questions/2213128/calculate-the-integral-int-e-sin-x-dx There are some good answers there. – Arun Bharadwaj Dec 07 '19 at 04:40

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As already mentioned, your work is wrong. Note that:

$$\ln\int f\ dx\ne\int\ln f\ dx$$

That is, you can't bring logarithms inside like that.

We have some Taylor expanding we can do:

$$e^{\sin(x)}=1+x+\frac12x^2-\frac18x^4-\frac1{15}x^5+\mathcal O(x^6)$$

Thus, we find that

$$\int e^{\sin(x)}\ dx=C+x+\frac12x^2+\frac16x^3-\frac1{40}x^5-\frac1{90}x^6+\mathcal O(x^7)$$

And I doubt of any closed form solutions.

Simply Beautiful Art
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