1. ## Integral help

$\int_{-2}^{2} \frac{x^2}{1+e^{sinx}} {d}x$

No hints given... Cheers for any help.

2. Here's the trick, let $u=-x$

You'll get: $I=
\int_{ - 2}^2 {\frac{{x^2 }}
{{1 + e^{\sin \left( x \right)} }}} dx = \int_{ - 2}^2 {\frac{{x^2 }}
{{1 + e^{ - \sin \left( x \right)} }}} dx
$

Thus: $
2 \cdot I= \int_{ - 2}^2 {\frac{{x^2 }}
{{1 + e^{\sin \left( x \right)} }}} dx + \int_{ - 2}^2 {\frac{{x^2 }}
{{1 + e^{ - \sin \left( x \right)} }}} dx
$

$
\int_{ - 2}^2 {\frac{{x^2 }}
{{1 + e^{\sin \left( x \right)} }}} dx + \int_{ - 2}^2 {\frac{{x^2 }}
{{1 + e^{ - \sin \left( x \right)} }}} dx = \int_{ - 2}^2 {x^2 \left( {\frac{1}
{{1 + e^{\sin \left( x \right)} }} + \frac{{e^{\sin \left( x \right)} }}
{{1 + e^{\sin \left( x \right)} }}} \right)} dx

$

Thus: $
2 \cdot I = \int_{ - 2}^2 {x^2 dx} = \frac{{2^4 }}
{3}\therefore I = \frac{8}
{3}
$

3. Wow thanks! Not at all obvious for me.

Here's another one I'm having problems with:

$
\int_{-\infty}^{\infty} e^{x/2}e^{-(x^2)/2} {d}x
$

4. Note that: $e^{-\frac{x^2-x}{2}}=e^{\frac{1}{8}}\cdot{e^{-\frac{(x-1/2)^2}{2}}}$

So: $\int_{-\infty}^{+\infty}e^{-\frac{x^2-x}{2}}dx=e^{\frac{1}{8}}\cdot{\int_{-\infty}^{+\infty}e^{-\left(\frac{x-1/2}{\sqrt[]{2}}\right)^2}dx}$

Now remember that: $\int_{-\infty}^{+\infty}e^{-x^2}dx=\sqrt{\pi}$ (the Gaussian Integral) and ...