>>891-893
∫1/{3+sin(x)^2}dx
=∫1/{3cos(x)^2+4sin(x)^2}dx
=∫(1/{3+4tan(x)^2})(1/cos(x)^2)dx
=∫(1/{3+4((√3/2)s)^2})(√3/2)ds ((√3/2)s = tan(x) とする)
=(1/(2√3))∫(1/{1+s^2})ds
=(1/(2√3))∫(1/{1+tan(z)^2})(1/cos(z)^2)dz (s = tan(z) とする)
=(1/(2√3))∫(1/{cos(z)^2+sin(z)^2})dz
=(1/(2√3))∫dz
=(1/(2√3))z +C
=(1/(2√3))arctan(s) +C
=(1/(2√3))arctan((2√3)tan(x)/3) +C

arctan((2√3)tan(x)/3) が x=±π/2で不連続であることに注意して定積分を求める
1/{3+sin(x)^2} は 周期π で周期的だから

∫[x=0〜π] 1/{3+sin(x)^2}dx
＝∫[x=-π/2〜π/2] 1/{3+sin(x)^2}dx
＝lim(w→π/2) {(1/2√3)arctan((2√3)tan(w)/3)-arctan((2√3)tan(-w)/3)}
＝(1/(2√3))(π/2-(-π/2))
＝π/(2√3) = 0.9068996821171...