cost2dt在0到x的积分

乐怎柄4698若函数f(x)在【0,1】上连续,证明∫f(sinx)=∫f(cosx) 0<x<π/2 -
贾纨叶18469887634 ______ 令x=π/2-t,则∫f(sinx)=∫f(cost) d( π/2-t) (t从π/2到0) =-∫f(cost) dt (t从π/2到0) =∫f(cost) dt (t从0到π/2) ==∫f(cosx) dx (x从0到π/2)

乐怎柄4698如何用换元法算根号根号(1 - X平方) X属于0到1的定积分 -
贾纨叶18469887634 ______ √(1-x^2)=√(1-sin^2t)=√cos^2t=cost

乐怎柄4698不定积分 取微分 得被积函数 定积分取微分 怎么算啊 -
贾纨叶18469887634 ______ =2xcos(x^2)dx 一般地 积分a(x)到b(x) f(t)dt的微分为 f(b(x))b'(x)dx-f(a(x))a'(x)dx

乐怎柄4698limx趋向0,∫上极限x下极限0costdt/x^2怎么做? -
贾纨叶18469887634 ______ (x->0) lim ∫ costdt / x^2 [t: 0, x] =(x->0) lim ∫ d(sint) / x^2 [t: 0, x] =(x->0) lim sint / x^2 [t: 0, x] =(x->0) lim (sinx-sin0) / x^2 =(x->0) lim sinx / x^2 =(x->0) lim sinx / x * (x->0)lim 1/x =1* (x->0)lim 1/x 是发散的,无极限

乐怎柄4698在积分区间【0,x】 求∫t*cost*dt=? -
贾纨叶18469887634 ______[答案] 原式=∫tdsint =tsint-∫sintdt =tsint+cost(0到x) =xsinx+cosx-0-cos0 =xsinx+cosx-1

乐怎柄4698第二类换元积分法 -
贾纨叶18469887634 ______ x=sint,dx=cost ∫x^2/(根号1-x^2=∫sin^2tdt=(1-cos2t)dt/2

乐怎柄4698∫2,0√(4 - x^2)dx怎么算 -
贾纨叶18469887634 ______[答案] 令x=2sint,dx=2costdt,x=2时,t= π/2,x=0时,t=0 ∫(2,0)√(4-x^2)dx =∫(π/2,0)2cost*2costdt =2∫(π/2,0)2cos²tdt =2∫(π/2,0)(1-cos2t)dt =2t-sin2t│(π/2,0) =π

乐怎柄4698若函数f(x)在【0,1】上连续,证明∫f(sinx)=∫f(cosx) 0 -
贾纨叶18469887634 ______[答案] 令x=π/2-t,则∫f(sinx)=∫f(cost) d( π/2-t) (t从π/2到0) =-∫f(cost) dt (t从π/2到0) =∫f(cost) dt (t从0到π/2) ==∫f(cosx) dx (x从0到π/2)

乐怎柄4698对根号下1 - x的平方,积分,区域为0到1 为什么答案是4分之派 -
贾纨叶18469887634 ______[答案] 令x=sint,则t∈[0,π/2],dx=costdt ∫【0→1】√(1-x²)dx =∫【0→π/2】cost ·costdt =∫【0→π/2】cos²tdt =∫【0→π/2】(1+cos2t)/2dt =[t/2+(sin2t)/4]【0→π/2】 =π/4+0-0-0 =π/4 答案:π/4

乐怎柄4698如何用换元法算根号根号(1 - X平方) X属于0到1的定积分设x=sint,dx=costdt,x=0,t=0,x=1,t=π/2,∫[0,1]√(1 - x^2)dx=∫[0,π/2]cost*costdt (根号怎么约去的 ) =∫[0,... -
贾纨叶18469887634 ______[答案] √(1-x^2)=√(1-sin^2t)=√cos^2t=cost