求不定积分 ∫1/2+sinx dx
2个回答
展开全部
令 tan(x/2) = t, 则 sint = 2t/(1+t^2), dx = 2dt/(1+t^2)
∫[1/(2+sinx)]dx = ∫{1/[2+ 2t/(1+t^2)]}2dt/(1+t^2)
= ∫dt/(t^2+t+1) = ∫dt/[(t+1/2)^2+3/4]
= (2/√3)arctan[(2t+1)/√3] + C
= (2/√3)arctan{[2tan(x/2)+1]/√3]} + C
∫[1/(2+sinx)]dx = ∫{1/[2+ 2t/(1+t^2)]}2dt/(1+t^2)
= ∫dt/(t^2+t+1) = ∫dt/[(t+1/2)^2+3/4]
= (2/√3)arctan[(2t+1)/√3] + C
= (2/√3)arctan{[2tan(x/2)+1]/√3]} + C
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
