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
# => -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#
