1. ## Difficult integrals

$\int\frac{xsinx}{1+(cosx)^{2}}dx$
$\int_{-\infty }^{\infty }e^{-x^{2}}dx$
$\int\frac{\sqrt{\sqrt{x^4 + 1}-x^{2}}}{x^4 + 1}dx$
$\int\frac{\sqrt{x}ln(x)e^{x}x^{3}}{\sqrt{lnx}}dx$
I have been working on these integral for 4 weeks... haven't been able to get anything yet. I am preparing for a competition and came across these bad boys.

2. ## Re: Difficult integrals

I found the solution to the second integral. It turned out to be the Gaussian Integral!

3. ## Re: Difficult integrals

I ve done for you the first one. Doesnt seem to have any sense but maybe I made a typing mistake.

http://integrals.wolfram.com/index.jsp?expr=(x*sin(x))%2F(1+%2B+cos(x)^2)&rando m=false

4. ## Re: Difficult integrals

Originally Posted by smokesalot
$\int\frac{xsinx}{1+(cosx)^{2}}dx$
$\int_{-\infty }^{\infty }e^{-x^{2}}dx$
$\int\frac{\sqrt{\sqrt{x^4 + 1}-x^{2}}}{x^4 + 1}dx$
$\int\frac{\sqrt{x}ln(x)e^{x}x^{3}}{\sqrt{lnx}}dx$
I have been working on these integral for 4 weeks... haven't been able to get anything yet. I am preparing for a competition and came across these bad boys.
$\displaystyle \int{\frac{x\sin{(x)}}{1 + \cos^2{(x)}}\,dx}$

Use integration by parts with $\displaystyle u = x \implies du = dx$ and $dv = \frac{\sin{(x)}}{1 + \cos^2{(x)}} \implies v = -\arctan{ \left[ \cos{(x)} \right] }$ and the integral becomes

$\displaystyle \int{\frac{x\sin{(x)}}{1 + \cos^2{(x)}}\,dx} = -x\arctan{\left[ \cos{(x)} \right]} + \int{\arctan{ \left[ \cos{(x)} \right]} \,dx}$

This does not have a closed-form solution in terms of elementary functions.

5. ## Re: Difficult integrals

wow nice, didn't see that! Thank you so much