1. ## Integral help

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

No hints given... Cheers for any help.

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

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

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

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

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

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

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

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

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

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

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