What is $f (x) = \int sec x - tan x d x$ $f(x) = int secx-tanx dx$ if $f (\frac{5 π}{3}) = 0$ $f((5pi)/3) = 0$ ?

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Answer 1 · 2016-09-22T20:23:31

$f (x) = ln ∣ ∣ 2 (2 + \sqrt{3}) (1 + sin x) ∣ ∣$ $f(x)=ln|2(2+sqrt3)(1+sinx)|$ .

$f (x) = \int (sec x - tan x) d x$ $f(x)=int(secx-tanx)dx$

$:. f(x)=intsecxdx-inttanxdx$

$=ln|secx+tanx|-ln|secx|$

$=ln|(secx+tanx)/secx|$

$=ln|secx/secx+tanx/secx|$

$=ln|1+sinx|+c$ .

To determine the const. $c", we use the given cond. : "f(5pi/3)=0$

$rArr ln|1+sin(5pi/3)|+c=0$

$"Since, "sin(5pi/3)=sin(2pi-pi/3)=-sin(pi/3)=-sqrt3/2$ , so,

$c=-ln|1-sqrt3/2|=-ln|(2-sqrt3)/2|=ln|2/(2-sqrt3)|=ln(2(2+sqrt3))$

$:. f(x)=ln|1+sinx|+ln(2(2+sqrt3))$ , or,

$f(x)=ln|2(2+sqrt3)(1+sinx)|$ .

Enjoy Maths.!

What is f(x) = int secx-tanx dxf(x)=∫secx−tanxdxf(x) = int secx-tanx dx if f((5pi)/3) = 0 f(5π3)=0f((5pi)/3) = 0 ?