What is #f(x) = int sin^2x-cotx dx# if #f((5pi)/4) = 0 #?

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Answer 1 · 2017-02-22T18:42:57

#f(x)=x/2-(sin2x)/4-ln|sinx|+1/4-5pi/8-1/2ln2.#

#f(x)=int(sin^2x-cotx)dx.#

#:. f(x)=intsin^2xdx-intcotxdx,#

#=int(1-cos2x)/2dx-ln|sinx|,#

#=1/2{int1dx-intcos2xdx}-ln|sinx|,#

#rArr f(x)=x/2-(sin2x)/4-ln|sinx|+C.#

#"To determine "C," we use the given cond. that "f(5pi/4)=0.#

#rArr 5pi/8-(sin(5pi/2))/4-ln|sin(5pi/4)|+C=0.#

#rArr 5pi/8-1/4-ln|-1/sqrt2|+C=0.#

#:. C=1/4-5pi/8-1/2ln2.#

Hence, #f(x)=x/2-(sin2x)/4-ln|sinx|+1/4-5pi/8-1/2ln2.#

Enjoy Maths.!