大师作文网第五题,求积分,
来源：学生作业帮 编辑：大师作文网作业帮 分类：数学作业 时间：2024/11/12 14:12:48
第五题,求积分,
∫cos2x/[(cosx)^2*(sinx)^2] *dx
=∫[(cosx)^2 - (sinx)^2]/[(cosx)^2 *(sinx)^2] *dx
=∫dx/(sinx)^2 - ∫dx/(cosx)^2
=∫(cscx)^2 *dx - ∫(secx)^2 *dx
=-cotx - tanx + C
∫dx/(1+sinx)
=∫dx/[sin(x/2) + cos(x/2)]^2
=∫[sec(x/2)]^2 *dx/[sec(x/2)*sin(x/2) + sec(x/2)*cos(x/2)]^2
=∫2* sec(x/2)*d(x/2) /[tan(x/2) + 1]^2
=2*∫d[tan(x/2)]/[tan(x/2) +1]^2
=2*∫[tan(x/2) +1]^(-2) *d[tan(x/2) +1]
=2* (-1) * [tan(x/2) +1]^(-1) + C
=-2/[tan(x/2) +1] + C
=∫[(cosx)^2 - (sinx)^2]/[(cosx)^2 *(sinx)^2] *dx
=∫dx/(sinx)^2 - ∫dx/(cosx)^2
=∫(cscx)^2 *dx - ∫(secx)^2 *dx
=-cotx - tanx + C
∫dx/(1+sinx)
=∫dx/[sin(x/2) + cos(x/2)]^2
=∫[sec(x/2)]^2 *dx/[sec(x/2)*sin(x/2) + sec(x/2)*cos(x/2)]^2
=∫2* sec(x/2)*d(x/2) /[tan(x/2) + 1]^2
=2*∫d[tan(x/2)]/[tan(x/2) +1]^2
=2*∫[tan(x/2) +1]^(-2) *d[tan(x/2) +1]
=2* (-1) * [tan(x/2) +1]^(-1) + C
=-2/[tan(x/2) +1] + C