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
I'll post up where I'm at with this. Perhaps you will see it before I do.
Let $\displaystyle t=\sin(x)$, $\displaystyle dt = \cos(x)dx$.
Integral becomes,
$\displaystyle \int \frac{e^{-t^2}t}{\cos(\arcsin(t))} dt = \int \frac{e^{-t^2}t}{\sqrt{1-t^2}} dt$.
I further simplified this by letting $\displaystyle y=t^2$, $\displaystyle dy = 2tdt$.
Then we get...
$\displaystyle \int \frac{e^{-t^2}t}{\sqrt{1-t^2}} dt = \frac{1}{2} \int \frac{e^{-y}}{\sqrt{1-y}} dy$.
However I can't advance past this yet! Perhaps you can.
$\displaystyle e^{-\sin^2(x)}\sin(x) = e^{-1 + \cos^2(x) }\sin(x) $
$\displaystyle = e^{-1} e^{\cos^2(x) }\sin(x) $
Sub. $\displaystyle \cos(x) = t ~~~ -\sin(x)~dx = dt$
It becomes $\displaystyle -e^{-1 }\int e^{t^2}~dt$
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 ...
Nice solution simplependulum.
This is what I further thought of at 4am!
Let $\displaystyle y=t^2 - 1$ => $\displaystyle t^2 = y+1$
$\displaystyle 2tdt = dy$
Integral becomes...
$\displaystyle \int \frac{e^{-t^2}t}{\sqrt{1-t^2}}dt = \frac{1}{2} \int \frac{e^{-y}e^{-1}}{\sqrt{-y}}
$
$\displaystyle = \frac{-ie^{-1}}{2} \int e^{-y}y^{-\tfrac{1}{2}} dy$
$\displaystyle = \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{y}\Big{)}$
$\displaystyle = \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{t^2 - 1}\Big{)}$
$\displaystyle = \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{\sin^2(x) - 1}\Big{)}$
$\displaystyle = \frac{-i e^{-1} \sqrt{\pi}}{2} erf \Big{(}\sqrt{1 - \cos^2(x) - 1}\Big{)}$
$\displaystyle = \frac{-i e^{-1} \sqrt{\pi}}{2} erf(i\cos(x))$
EDIT: Apparently that minus should not be there (or there should be a minus in front of the $\displaystyle i\cos(x)$) but I can't figure out what I've done wrong after skimming over it.