dx/(x+根号下(1-x平方))的不定积分
问题描述:
dx/(x+根号下(1-x平方))的不定积分
答
令x=siny,则:√(1-x^2)=√[1-(siny)^2]=cosy, y=arcsinx, dx=cosydy.原式=∫[cosy/(siny+cosy)]dy =∫{cosy(cosy-siny)/[(cosy)^2-(siny)^2]}dy =∫[(cosy)^2/cos2y]...