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

1 Answer
Feb 22, 2017

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

Explanation:

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.!