`p=3int(sin\ pi t)/(2+cos\ pi t)dt`

Put `u = 2 + cos\ π t`, then

`du = −π\ sin\ π t\ dt`

So

`p=3int(sin\ pi t)/(2+cos\ pi t)dt`

`=-3/piint(-pi\ sin\ pi t)/(2+cos\ pi t)dt`

`=-3/pi[ln|2+cos\ pi t|]+K`