Question

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

Answer 1

`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`