As it stands, #intsin^2xdx# has no clear solutions. This is one of those cases where knowing your algebra and trig comes in real handy.
You may recall that #sin^2x=1/2(1-cos2x)# (if you don't, that's fine because I forgot too). So, instead of integrating #sin^2x#, we can integrate #1/2(1-cos2x)# and see if that gets us anywhere:
#int1/2(1-cos2x)dx#
#=1/2int1-cos2xdx#
We can use the sum rule to split this into:
#1/2int1dx-1/2intcos2xdx#
We're getting somewhere because the first of these integrals, #int1dx#, is the "perfect integral" and evaluates to #x#:
#1/2x-1/2intcos2xdx#
We just have to worry about #intcos2xdx#. Let's start off with what we know:#intcosxdx=sinx# because the derivative of #sinx# is #cosx#. We just have to adjust for that pesky #2#.
Let's think for a moment. #intcos2xdx# essentially means that if we take the derivative of our solution, we should get #cos2x#. Let's guess a solution of #1/2sin2x# and see what happens when we differentiate it:
#d/dx1/2sin2x=1/2*(2x)'*cos2x=1/2*2*cos2x=cos2x#
Since that solution checks out, we can say that #intcos2xdx=1/2sin2x#
Ok, so we have #1/2x-1/2intcos2xdx#. We just found the last integral, so our solution is #1/2x-1/2(1/2sin2x)+C=1/2x-1/4sin2x+C# (don't forget the integration constant #C#!)
