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

Question

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

Luke Alvarez · Accepted Answer

#f(x) = lnabs(sinx) -1/2 ln abs(secx+tanx) -1 -1/2 ln3+ln2 +1/2 ln (2 +sqrt3)#

#f(x) = -1+int_(pi/3)^x (cot t -sec 2t) dt#

using the linearity of the integral:

#f(x) = -1+int_(pi/3)^x cot t dt -int_(pi/3)^x sec 2t dt#

#f(x) = -1+int_(pi/3)^x cost/sint dt -1/2 int_(pi/3)^x sec 2t d(2t)#

#f(x) = -1+int_(pi/3)^x (d(sint))/sint dt -1/2 int_(pi/3)^x sec 2t d(2t)#

#f(x) = -1+ lnabs(sinx)-ln abs sin(pi/3) -1/2 ln abs(sec2x+tan2x) +1/2 ln abs(sec((2pi)/3) + tan((2pi)/3))#

#f(x) = lnabs(sinx) -1/2 ln abs(secx+tanx) -1 -ln (sqrt3/2 )+1/2 ln abs(-2 -sqrt3)#

#f(x) = lnabs(sinx) -1/2 ln abs(secx+tanx) -1 -1/2 ln3+ln2 +1/2 ln (2 +sqrt3)#

Charles Anctil · Answer

undefined To find $ f(x) = \int \cot(x) - \sec^2(x) \, dx $ given that $ f\left(\frac{\pi}{3}\right) = -1 $, we integrate the function and then use the given information to solve for the constant of integration.

The integral of $ \cot(x) - \sec^2(x) $ is $ -\ln|\sin(x)| - \tan(x) + C $, where $ C $ is the constant of integration.

Given $ f\left(\frac{\pi}{3}\right) = -1 $, we substitute $ x = \frac{\pi}{3} $ into the expression for $ f(x) $:

$$
-1 = -\ln|\sin\left(\frac{\pi}{3}\right)| - \tan\left(\frac{\pi}{3}\right) + C
$$

$$
-1 = -\ln\left|\frac{\sqrt{3}}{2}\right| - \sqrt{3} + C
$$

$$
C = -1 + \sqrt{3} + \ln\left|\frac{\sqrt{3}}{2}\right|
$$

Thus, the function $ f(x) $ is:

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