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
Since we know $\displaystyle \displaystyle 1+tan^2(\theta)=sec^2(\theta)$, let $\displaystyle \displaystyle u=tan(\theta) \ \mbox{then} \ du=sec^2(\theta)d\theta$
Now by subing, we obtain:
$\displaystyle \displaystyle \int\frac{sec^2(\theta)d\theta}{\sqrt{1+tan^2(\the ta)}}$
$\displaystyle =\displaystyle \int\frac{sec^2(\theta)d\theta}{\sqrt{sec^2(\theta )}}$
$\displaystyle =\displaystyle \int\frac{sec^2(\theta)d\theta}{sec(\theta)}$
$\displaystyle =\displaystyle \int sec(\theta)d\theta$
Then integrate sec and remember $\displaystyle \displaystyle\theta=tan^{-1}(u) \ \mbox{and} \ u=ln(x)$