Question

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

Answer 1

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

Explanation:

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`