∫sinπrrdr

To evaluate the integral ∫sin(πr)rdr, we can use integration by parts.

Let's choose u = r and dv = sin(πr)dr.

Then, du = dr and v = -1/πcos(πr).

Using the integration by parts formula, we have:

∫sin(πr)rdr = uv - ∫vdu = -1/πrcos(πr) - ∫(-1/πcos(πr))dr = -1/πrcos(πr) + 1/π∫cos(πr)dr = -1/πrcos(πr) + 1/π(1/πsin(πr)) = -1/πrcos(πr) + 1/π^2sin(πr) + C

So, the integral of sin(πr)rdr is equal to -1/πrcos(πr) + 1/π^2sin(πr) + C, where C is the constant of integration