# Integration problem

• Jun 9th 2010, 12:18 PM
simonwei
Integration problem
Hi, dear all:

I'm wondering how to integrate (exp(-sin(x)^2))*sin(x) with respect to x? I believe it has something to do with error function, but just can't figure out a right form. Pleas help me! Thank you very much.

simonwei
• Jun 9th 2010, 02:13 PM
I'll post up where I'm at with this. Perhaps you will see it before I do.

Let $t=\sin(x)$, $dt = \cos(x)dx$.

Integral becomes,

$\int \frac{e^{-t^2}t}{\cos(\arcsin(t))} dt = \int \frac{e^{-t^2}t}{\sqrt{1-t^2}} dt$.

I further simplified this by letting $y=t^2$, $dy = 2tdt$.

Then we get...

$\int \frac{e^{-t^2}t}{\sqrt{1-t^2}} dt = \frac{1}{2} \int \frac{e^{-y}}{\sqrt{1-y}} dy$.

However I can't advance past this yet! Perhaps you can.
• Jun 9th 2010, 08:46 PM
simplependulum
$e^{-\sin^2(x)}\sin(x) = e^{-1 + \cos^2(x) }\sin(x)$

$= e^{-1} e^{\cos^2(x) }\sin(x)$

Sub. $\cos(x) = t ~~~ -\sin(x)~dx = dt$

It becomes $-e^{-1 }\int e^{t^2}~dt$

It is not the error function but it is close , seemingly no one can use a finite number of elementary functions to express this integral ...
• Jun 10th 2010, 05:20 AM
Nice solution simplependulum.

This is what I further thought of at 4am!

Let $y=t^2 - 1$ => $t^2 = y+1$

$2tdt = dy$

Integral becomes...

$\int \frac{e^{-t^2}t}{\sqrt{1-t^2}}dt = \frac{1}{2} \int \frac{e^{-y}e^{-1}}{\sqrt{-y}}
$

$= \frac{-ie^{-1}}{2} \int e^{-y}y^{-\tfrac{1}{2}} dy$

$= \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{y}\Big{)}$

$= \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{t^2 - 1}\Big{)}$

$= \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{\sin^2(x) - 1}\Big{)}$

$= \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{1 - \cos^2(x) - 1}\Big{)}$

$= \frac{-i e^{-1} \sqrt{\pi}}{2} erf(i\cos(x))$

EDIT: Apparently that minus should not be there (or there should be a minus in front of the $i\cos(x)$) but I can't figure out what I've done wrong after skimming over it.