# What is f(x) = int cotx-sec2x dx if f(pi/3)=-1 ?

Jun 12, 2018

$f \left(x\right) = \ln \left\mid \sin x \right\mid - \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid - 1 - \frac{1}{2} \ln 3 + \ln 2 + \frac{1}{2} \ln \left(2 + \sqrt{3}\right)$

#### Explanation:

$f \left(x\right) = - 1 + {\int}_{\frac{\pi}{3}}^{x} \left(\cot t - \sec 2 t\right) \mathrm{dt}$

using the linearity of the integral:

$f \left(x\right) = - 1 + {\int}_{\frac{\pi}{3}}^{x} \cot t \mathrm{dt} - {\int}_{\frac{\pi}{3}}^{x} \sec 2 t \mathrm{dt}$

$f \left(x\right) = - 1 + {\int}_{\frac{\pi}{3}}^{x} \cos \frac{t}{\sin} t \mathrm{dt} - \frac{1}{2} {\int}_{\frac{\pi}{3}}^{x} \sec 2 t d \left(2 t\right)$

$f \left(x\right) = - 1 + {\int}_{\frac{\pi}{3}}^{x} \frac{d \left(\sin t\right)}{\sin} t \mathrm{dt} - \frac{1}{2} {\int}_{\frac{\pi}{3}}^{x} \sec 2 t d \left(2 t\right)$

$f \left(x\right) = - 1 + \ln \left\mid \sin x \right\mid - \ln \left\mid \sin \right\mid \left(\frac{\pi}{3}\right) - \frac{1}{2} \ln \left\mid \sec 2 x + \tan 2 x \right\mid + \frac{1}{2} \ln \left\mid \sec \left(\frac{2 \pi}{3}\right) + \tan \left(\frac{2 \pi}{3}\right) \right\mid$

$f \left(x\right) = \ln \left\mid \sin x \right\mid - \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid - 1 - \ln \left(\frac{\sqrt{3}}{2}\right) + \frac{1}{2} \ln \left\mid - 2 - \sqrt{3} \right\mid$

$f \left(x\right) = \ln \left\mid \sin x \right\mid - \frac{1}{2} \ln \left\mid \sec x + \tan x \right\mid - 1 - \frac{1}{2} \ln 3 + \ln 2 + \frac{1}{2} \ln \left(2 + \sqrt{3}\right)$