∫dx/[a*sin^2(x-π/2)+b*sin(x-π/2)+c] ∫dx/[a*sin^2(x-π/2)+b*sin(x-π/2)+c] X从0到2π的积分值a,b,c,为常数

∫cosx/(a^2+sin^2x)dx ∫x/sin^2(x) dx ∫sin(x) cos^2(x)dx 定积分 从b 到 a 求 d∫ sin(x^2)dx/dx ∫sin^2x/(1+sin^2x )dx求解, ∫sin(6x)sin(2x)dx=? ∫(2/sin^2x)dx= ∫(1-sin^2( x/2))dx 求∫ sin^2 x/2 dx .. 求不定积分：∫sin(x^2)dx 求不定积分∫sin(x/2)dx ∫[(sin^2)x]dx求不定积分 ∫(cosx／1+sin^2x)dx ∫sinx/(cosx-sin^2x)dx ∫sin(√2x)dx ∫ 4/sin^2x dx ∫sin^2x(1+tanx)dx 求∫sin(x^2)dx