0128132人目の素数さん
2019/09/20(金) 16:42:41.56ID:lq2/XErocos(π/2) = cos(3*π/2) = 0
なので、少し面倒ですね。
∫_{0}^{2*π} 1 / (a^2 * cos^2(x) + b^2 * sin^2(x)) dx
=
∫_{-π/2}^{3*π/2} 1 / (b^2 * cos^2(x) + a^2 * sin^2(x)) dx
=
∫_{-π/2}^{π/2} 1 / (b^2 * cos^2(x) + a^2 * sin^2(x)) dx
+
∫_{π/2}^{3*π/2} 1 / (b^2 * cos^2(x) + a^2 * sin^2(x)) dx