<p>How do you integrate (sin x)^2 (also written as sin^2 x)?

Thanks in advance. :)</p>

<p>(x-sinxcosx)/2.. do it by parts u = sinx dv = sinxdx.. ithink anyway</p>

<p>Ok... you use the half angle formula- the formula is for sin of (1/2)X, But it equals the square root of ((1/2)(1-cosx)) Just change it so that sin squared x equals (1/2)(cos 2x) Are you doing integration by parts? becuase f so you will than do integral and get sin(2x) where you change it to 2sinxcosx. Hope I helped!</p>

<p>can anyone confirm this?</p>

<p>Roh kay, I did it differently, but it's still right, i just busted out an identity cos2x = 1-2sin^2(x). Solve for sin^2(x) to get:</p>

<p>[Integral] cos2x-1/-2, then pull out the -1/2 and split it up.</p>

<p>-1/2 [ (int.cos2xdx) - x] remember integral of 1dx is x.</p>

<p>Using u substitution u = 2x, du/2=dx, so cosu/2 du, then pull out 1/2 to get integral of cos u which is sin u, put 2x back in then and multiply the 1/2 we pulled out to get sin2x/2.</p>

<p>-1/2 [sin2x/2 - x] = x/2 - sin2x/4</p>

<p>Ahhhh I love math.</p>

<p>Oh yeah and I forgot........</p>

<p>x/2 - sin2x/4 + C (Never forget the C; I learned the hard way. The God of Constants becomes angry and floods your whole town with a monster slope field of water.)</p>

<p>You can use the reduction formula. </p>

<p>which means sin^2(x)=-1/2(sin(x)cos(x)+1/2 int.1 dx

= -1/2(sin(x)cos(x)) +1/2(x)+c</p>