How do I integrate sin^{2} x

First convert sin-squared-x to its double-angle identity: 1/2[1-cos(2x)] Then integrate that expression by conventional methods to obtain the answer: x/2 - sin(2x)/4

sin ^2 x = (1 - cos2x) / 2

Thanks

And cos^{2} x =1/2(cos(2x)+1) is that right

yepp thats right

Thanks

:):)

hi eliza, they may have forgotten the integral of dx=x so integral of sin^{2} x dx =integral of (1/2)(1-cos2x)dx = (1/2)(x-sin2x) +C

integral of sin^{2} x dx =integral of (1/2)(1-cos2x)dx = (1/2)(x-sin2x) +C answer....

did you get it eliza?

therefore integral of sin^{2} x dx =integral of (1/2)(1-cos2x)dx = (1/2)(x-sin2x) +C or =(1/2)(x-2sinxcosx)+C or =(x/2 - sinxcosx)+C ans....

Thanks yeah sorry I know how to integrate by parts I was just being silly and forgetting the other identity