∫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

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∫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
∫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

∫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
LZ好,∫(sin5x)(sin3x)dx
=∫1/2[cos(5x - 3x) - cos(5x + 3x)]dx
=1/2 ∫[cos2x - cos8x]dx
=1/2∫cos2xdx-1/2∫cos8xdx
=1/2*1/2∫cos2xd2x-1/2*1/8∫cos8xd8x
=1/4*sin2x-1/16*sin8x+C
C为任意常数
利用到了:sina*sinb=1/2[cos(a-b)-cos(a+b)]
和积分公式:∫cosxdx=sinx+C 12475希望对你有帮助!

∫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