Question

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

Answer 1

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

Explanation:

`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)`