# Math Help - weird integral

1. ## weird integral

so i have 2pi ∫ (a=1 b= infiniti) 1/x sqrt of 1+ (lnx)^2

how do i start this beast? trig sub was recommended to me, but i cant see how to start that

2. I think u sub with u being the ln(x) since du would equal $\frac{1}{x}$ and then trig sub where u=tan

3. but can you still use trig sub on? ∫du sqrt 1+u^2

4. Yes, $\displaystyle 1+tan^2(\theta)=sec^2(\theta)$

5. oh no im sorry i mean triangle trig, where its like du/sqrt 1+u^2 and u make a triangle with the sqrt on the hyp, u on the tall side and 1 on the base. then find that sec theta = sqrt 1+u^2 tan theta = u sec ^2 theta dtheta = du

6. Since we know $\displaystyle 1+tan^2(\theta)=sec^2(\theta)$, let $\displaystyle u=tan(\theta) \ \mbox{then} \ du=sec^2(\theta)d\theta$

Now by subing, we obtain:

$\displaystyle \int\frac{sec^2(\theta)d\theta}{\sqrt{1+tan^2(\the ta)}}$

$=\displaystyle \int\frac{sec^2(\theta)d\theta}{\sqrt{sec^2(\theta )}}$

$=\displaystyle \int\frac{sec^2(\theta)d\theta}{sec(\theta)}$

$=\displaystyle \int sec(\theta)d\theta$

Then integrate sec and remember $\displaystyle\theta=tan^{-1}(u) \ \mbox{and} \ u=ln(x)$

7. how did you just let tan theta = u?

8. We have 1+u^2 and 1+tan^2=sec^2. By making that substitution, we have a well known trig identity. When we make u=tan and substitute, we have 1+(tan)^2=1+tan^2=sec^2. Then by taking the square root, we get sec.

9. i dont understand how you can just say that u, which = ln|x| can all of a sudden = tan theta

10. You'll probably need to open your calculus book/notes and review integration all over again.

11. When integrating, we can do as many substitutions as it takes just like we can use L'Hopital's Rule more than once too.