sinx^6/(sinx^6+cosx^6)不定积分?
1个回答
展开全部
目测本题应该是0→π/2的定积分.,1,
瞬间陌尘 举报
额。如果是定积分怎么做? 用一个结论:∫[0→π/2] f(sinx) dx = ∫[0→π/2] f(cosx) dx 因此: ∫[0→π/2] (sinx)^6/[(sinx)^6+(cosx)^6] dx =∫[0→π/2] (cosx)^6/[(sinx)^6+(cosx)^6] dx =(1/2){ ∫[0→π/2] (sinx)^6/[(sinx)^6+(cosx)^6] dx + ∫[0→π/2] (cosx)^6/[(sinx)^6+(cosx)^6] dx } =(1/2)∫[0→π/2] 1 dx =π/4 定积分与不定积分的差异是很大的,不能随便把定积分的题当作不定积分来问。,次方在里面还是外面?究竟是sin^6(x)还是sin(x^6)?是(sinx)^6令A=∫sin^6(x)/(sin^6(x)+cos^6(x))dx,B=∫cos^6(x)/(sin^6(x)+cos^6(x))dx, 则A+B=∫dx=x+C, A-B=∫(sin^2(x)-cos^2(x))*(sin^4(x)+sin^2(x)cos^2(x)+cos^4(x))/(sin^2(x)+c...,1,
瞬间陌尘 举报
额。如果是定积分怎么做? 用一个结论:∫[0→π/2] f(sinx) dx = ∫[0→π/2] f(cosx) dx 因此: ∫[0→π/2] (sinx)^6/[(sinx)^6+(cosx)^6] dx =∫[0→π/2] (cosx)^6/[(sinx)^6+(cosx)^6] dx =(1/2){ ∫[0→π/2] (sinx)^6/[(sinx)^6+(cosx)^6] dx + ∫[0→π/2] (cosx)^6/[(sinx)^6+(cosx)^6] dx } =(1/2)∫[0→π/2] 1 dx =π/4 定积分与不定积分的差异是很大的,不能随便把定积分的题当作不定积分来问。,次方在里面还是外面?究竟是sin^6(x)还是sin(x^6)?是(sinx)^6令A=∫sin^6(x)/(sin^6(x)+cos^6(x))dx,B=∫cos^6(x)/(sin^6(x)+cos^6(x))dx, 则A+B=∫dx=x+C, A-B=∫(sin^2(x)-cos^2(x))*(sin^4(x)+sin^2(x)cos^2(x)+cos^4(x))/(sin^2(x)+c...,1,
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