Question

how do I find the indefinite integral of pi *sin pi x dx

Answer 1

\[\pi\int\sin(\pi x)dx\] use \(u=\pi x\), \(du=\pi dx\) and so \(\frac{1}{\pi}du=dx\)
gives \(\int \sin(u)du\) and now it should be easier

Answer 2

This is cal 2 I'm assuming?

∫ πsin(πx) dx let u = πx, then du= π dx, so dx = 1/π du
=1/π * ∫ πsin(u)du
=∫sinu du
=cosu
=cos(πx)