∫sec²xdx/(2+tg²x)
=∫dtgx/(2+tg²x)
=∫√2d(tgx/√2)/2[1+(tgx/√2)²]
=√2/2*arctan(tanx/√2)+C
所以原式=√2/2*[arctan(tanπ/√2)-arctan(tan0/√2)]=0