Using the result: ∫(2xsin(x)cos(x))dx = -1⁄2[xcos(2x)-1⁄2sin(2x)] calculate ∫sin²(x) dx using integration by parts

Recall that ∫uv'=uv- ∫u'v Set u=sin²(x), v'=1 Therefore u'=2sin(x)cos(x) and v=x which gives us the following:

∫sin²(x)dx = xsin²(x) - ∫2xsin(x)cos(x)dx

The second integral in the above expression is given in the question so we then have the form:

∫sin²(x)dx = xsin²(x) +1⁄2[xcos(2x)-1⁄2sin(2x)]

which can be rearranged to give:

∫sin²(x)dx = 1⁄2x[2sin²(x) + cos(2x)] - 1⁄4sin(2x) + c

We can then employ the identity: cos(2x) = cos²(x) - sin²(x) to give us:

∫sin²(x)dx = 1⁄2x[sin²(x) + cos²(x)] - 1⁄4sin(2x) + c

Finally the identity: sin²(x) + cos²(x) = 1 is used to produce:

∫sin²(x)dx = 1⁄2x- 1⁄4sin(2x) + c

where c is the constant of integration