What is #f(x) = int secx-tanx dx# if #f((5pi)/3) = 0 #?

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Answer 1 · 2016-09-22T20:23:31

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

#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.!