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

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Answer 1 · 2016-06-30T13:40:53

#f(x) = 1/4sin(2x) - cos(x) - x/2 + 1 + (3sqrt(3))/8 + pi/12#

The first term is simple.

#int sin(x) dx = -cos(x) + C_1# so we now look at the second term;

# - int sin^2(x) dx = - 1/2*int (1 - cos(2x))dx#

#= -1/2*(x - 1/2sin(2x)) + C_2#

Combining these we obtain that

#f(x) = 1/4sin(2x) - cos(x) - x/2 + C ( = C_1 + C_2)#

#f(pi/6) = 1 = 1/4*sin(pi/3) - cos(pi/6) - pi/12 + C#

#1 = (sqrt(3))/8 - (sqrt(3))/2 - pi/12 + C#

#C = 1 - (sqrt(3))/8 + (sqrt(3))/2 + pi/12 = 1 + (3sqrt(3))/8 + pi/12#

#f(x) = 1/4sin(2x) - cos(x) - x/2 + 1 + (3sqrt(3))/8 + pi/12#