题目

求∫sec³xdx,

提问时间：2021-01-18

答案

原式=∫dx/cos³x
=∫cosxdx/(cos²x)²
=∫d(sinx)/(1-sin²x)²
=(1/4)∫[1/(1+sinx)+1/(1-sinx)+1/(1+sinx)²+1/(1-sinx)²]d(sinx)
=(1/4)[ln│1+sinx│-ln│1-sinx│-1/(1+sinx)+1/(1-sinx)]+C (C是积分常数)
=(1/4)[ln│(1+sinx)/(1-sinx)│+2sinx/(1-sin²x)]+C
=(1/4)[ln│(1+sinx)²/(1-sin²x)│+2sinx/(1-sin²x)]+C
=(1/4)[2ln│(1+sinx)/cosx│+2sinx/cos²x]+C
=(1/2)[ln│(1+sinx)/cosx│+sinx/cos²x]+C
=(1/2)(ln│secx+tanx│+secxtanx)+C.