What is f(x) = int cosxsinx-sin^2x dx if f(pi/6)=1 ?

1 Answer
Apr 15, 2017

f(x) = -1/4cos2x+1/4sin2x -1/2x + 9/8-sqrt(3)/8 +pi/12

Explanation:

Weh have:

f(x) = int \ cosxsinx-sin^2x \ dx

We can use the identities:

sin2A -= 2sinAcosA
cos2A -= 1-2sin^2A

Which if applied to the integrand gives us:

f(x) = int \ 1/2sin2x +1/2(cos2x-1) \ dx
\ \ \ \ \ \ \ = -1/4cos2x+1/4sin2x -1/2x + C

We are given that f(pi/6)=1

=> -1/4cos(pi/3)+1/4sin(pi/3) -1/2*pi/6 + C = 1
:. -1/4*1/2+1/4*sqrt(3)/2 -pi/8 + C = 1
:. -1/8+sqrt(3)/8 -pi/13 + C = 1
:. C=9/8-sqrt(3)/8 +pi/12

Leading to the solution:

f(x) = -1/4cos2x+1/4sin2x -1/2x + 9/8-sqrt(3)/8 +pi/12