这题怎么做?求不定积分
3个回答
展开全部
∫ dx/(3sinx+4cosx +5)
=(1/5)∫ dx/[(3/5)sinx+(4/5)cosx +1 ]
=(1/5)∫ dx/[cos(x-φ) +1]
where cosφ = 4/5 , sinφ = 3/5
=(1/5)∫ dx/ { 2(cos[(x-φ)/2] )^2}
=(1/10)∫ { sec[(x-φ)/2] }^2 dx
=(1/5) tan[(x-φ)/2] + C
=(1/5)∫ dx/[(3/5)sinx+(4/5)cosx +1 ]
=(1/5)∫ dx/[cos(x-φ) +1]
where cosφ = 4/5 , sinφ = 3/5
=(1/5)∫ dx/ { 2(cos[(x-φ)/2] )^2}
=(1/10)∫ { sec[(x-φ)/2] }^2 dx
=(1/5) tan[(x-φ)/2] + C
本回答被网友采纳
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