Question about integrals in calculus

How do you find the integral of (sinx)^2 ?

I used integration by parts, but when I do integration by parts a second time my terms all cancel themselves out. Please help! :/

the equation for integration by parts is

u*dv = u*v - the integral of (v*du)

cos(2x)=1-2(sin(x))^2, then (sin(x))^2=(1-cos(2x))/2, from that you get that

integral[(sin(x))^2] = integral[(1-cos(2x))/2]

Integrate "1" and "cos2x" independently..
integration of 1 with respect to x is = "x"
integration of cos2x with respect to x is ="(sin2x)/2"
so the answer is =1/2[x-(sin2x/2)]+c

=½ x − ¼ sin 2x + C

∫sin² x dx = ½ x − ¼ sin 2x + C

using the trigonometric identity:
(sinx)^2 = (1-cos2x)/2
then,
let S = integral sign
S(sinx)^2 = S(1-cos2x)/2 dx
= S(1/2)dx - S(cos2x/2)dx

change of variables:
let u = 2x
du/dx = 2
du = 2dx
du/2 = dx

= (1/2)x - (1/2)S(cosu)(du/2)
= (1/2)x - (1/4)S(cosu)du
= (1/2)x - (1/4)[sinu] + C

since u = 2x

= (1/2)x - (1/4)[sin2x] + C
thus,
S(sinx)^2 = x/2 - sin2x/4 + C

Use the identity: (sinx)^2 = 1/2-cos(2x)/2 and then integrate.