# How do you evaluate the integral int 1/xcos(lnx)dx?

Feb 14, 2017

$\int \frac{1}{x} \cos \left(\ln x\right) \mathrm{dx} = \sin \left(\ln x\right) + C$

#### Explanation:

Substitute:

$t = \ln x$
$\mathrm{dt} = \frac{\mathrm{dx}}{x}$

so that:

$\int \frac{1}{x} \cos \left(\ln x\right) \mathrm{dx} = \int \cos t \mathrm{dt} = \sin t + C$

Undo the substitution and we have:

$\int \frac{1}{x} \cos \left(\ln x\right) \mathrm{dx} = \sin \left(\ln x\right) + C$

Feb 14, 2017

The integral equals $\sin \left(\ln x\right) + C$

#### Explanation:

This is a substitution problem. Note that $\frac{d}{\mathrm{dx}} \ln x = \frac{1}{x}$, so if we let $u = \ln x$, then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $\mathrm{dx} = x \mathrm{du}$.

$\implies \int \frac{1}{x} \cos \left(u\right) \cdot x \mathrm{du}$

$\implies \int \cos u \mathrm{du}$

$\implies \sin u + C$

$\implies \sin \left(\ln x\right) + C$

Hopefully this helps!