∫ x^5*√1-x²dx=？ ∫ sin³xcos³xdx=? ∫ √（a²-x²）dx=?

Question

如题所述

Answer 1

(1)
∫x^5.√(1-x^2)dx
let
x=siny
dx=cosy dy
∫x^5.√(1-x^2)dx
=∫ (siny)^5 .(cosy)^2 dy
=-∫ (1 -(cosy)^2)^2. (cosy)^2 dcosy
=-∫[ (cosy)^2 - 2(cosy)^4 + (cosy)^6] dcosy
=-(1/3)(cosy)^3 + (2/5)(cosy)^5 -(1/7)(cosy)^7 + C
=-(1/3)(1-x^2)^(3/2) + (2/5)(1-x^2)^(5/2) -(1/7)(1-x^2))^(7/2) + C

(2)
∫ (sinx)^3. (cosx)^2dx
=-∫ [1-(cosx)^2]. (cosx)^2dcosx
= -(1/3)(cosx)^3 + (1/5)(cosx)^5 + C

(3)
∫√(a^2-x^2）dx
let
x=asiny
dx=acosy dy
∫√(a^2-x^2）dx
=a^2∫ (cosy)^2 dy
=(a^2/2)∫ (1+cos2y) dy
=(a^2/2)[ y+(1/2)sin2y ] + C
=(a^2/2)[ arcsin(x/a)+ x√(a^2-x^2）/a^2 ] + C追问

怎样才能学好微积分？

Answer 2

提问不清楚，无法判断，无法回答问题，请收回。
这类型的题，以后还是不要分拣进来的好，对答题者没有任何途径回答。追问

怎么不清楚了，明明是你自己不知道做

Answer 3

(1)
∫x^5.√(1-x^2)dx
let
x=siny
dx=cosy dy
∫x^5.√(1-x^2)dx
=∫ (siny)^5 .(cosy)^2 dy
=-∫ (1 -(cosy)^2)^2. (cosy)^2 dcosy
=-∫[ (cosy)^2 - 2(cosy)^4 + (cosy)^6] dcosy
=-(1/3)(cosy)^3 + (2/5)(cosy)^5 -(1/7)(cosy)^7 + C
=-(1/3)(1-x^2)^(3/2) + (2/5)(1-x^2)^(5/2) -(1/7)(1-x^2))^(7/2) + C
(2)
∫ (sinx)^3. (cosx)^2dx
=-∫ [1-(cosx)^2]. (cosx)^2dcosx
= -(1/3)(cosx)^3 + (1/5)(cosx)^5 + C
(3)
∫√(a^2-x^2）dx
let
x=asiny
dx=acosy dy
∫√(a^2-x^2）dx
=a^2∫ (cosy)^2 dy
=(a^2/2)∫ (1+cos2y) dy
=(a^2/2)[ y+(1/2)sin2y ] + C
=(a^2/2)[ arcsin(x/a)+ x√(a^2-x^2）/a^2 ] + C

Answer 4

后两个不说了，比较简单，就说第一个吧，当算到siny∧5cosy∧2时，把cosy²写成1-siny²然后二者相乘，∫（sin∧5y-sin∧7y）dy，这样子好算