Question

find the integration of sin^2 x dx=???

Answer 1

Let, I = int [(x^2)sinx dx]

u = x^2, du = 2x dx
dv = sinx dx, v = -cosx

I = -(x^2)cosx - int [-2xcosx dx]
I = -(x^2)cosx + int [2xcosx dx]

Again integrate the second part by parts,
int [2xcosx dx]
u = 2x
du = 2 dx
dv = cosx dx
v = sinx
int [2xcosx dx] = 2xsinx - int [2sinx dx]
int [2xcosx dx] = 2xsinx + 2cosx +c

I = -(x^2)cosx + 2xsinx + 2cosx + c

Answer 2

∫ sin²x dx

since cos2x = 1 - 2sin²x
sin²x = ½ - ½cos2x

∫( ½ - ½cos2x )

= ½x -¼sin2x + C

@qamy

Answer 3

@qamy Your integration is complete.