Recently, we review inverse functions and how to compute derivatives of inverse functions, especially the inverse trigonometric functions and their derivatives.
Today we firstly review the chain rule (the name of the method to compute derivatives of composite of functions.)
If 
and 
, then 
. So we can regard 
#card=math&code=y%3Dy%28x%29&id=FU6TH) as a function of 
. Thus we can compute its derivative with respect to 
: 
%20%3D%20y’(u)%7C%7Bu%20%3D%20u(x)%7D%20%5Ccdot%20u’(x)#card=math&code=y%27%28x%29%20%3D%20y%27%28u%29%7C%7Bu%20%3D%20u%28x%29%7D%20%5Ccdot%20u%27%28x%29&id=EVi9f)
Example: Find the derivative of 
%20%3D%20x%20%5B(2x%2B3)%5E3%5D%5E%7B-1%7D%20%3D%20x%20%5Ccdot%20(2x%2B3)%5E%7B-3%7D#card=math&code=h%28x%29%20%3D%20x%20%5B%282x%2B3%29%5E3%5D%5E%7B-1%7D%20%3D%20x%20%5Ccdot%20%282x%2B3%29%5E%7B-3%7D&id=YSJHk)
Solution: Write 
%5E%7B-3%7D%20%3D%20u%5E%7B-3%7D#card=math&code=%282x%2B3%29%5E%7B-3%7D%20%3D%20u%5E%7B-3%7D&id=WjOBS) with 
.
%3Dx%5E%7B-3%7D#card=math&code=f%28x%29%3Dx%5E%7B-3%7D&id=oR6hp), 
%3D2x%2B3#card=math&code=g%28x%29%3D2x%2B3&id=BAnjn), so 
)%3D(2x%2B3)%5E%7B-3%7D#card=math&code=f%28g%28x%29%29%3D%282x%2B3%29%5E%7B-3%7D&id=ofWEH) .
Now apply the chain rule:
%3D%20-3%20x%5E%7B-4%7D#card=math&code=f%27%28x%29%3D%20-3%20x%5E%7B-4%7D&id=Jrzyp) , then evaluating at #card=math&code=g%28x%29&id=z9tbn): 
%5E%7B-4%7D#card=math&code=-3%282x%2B3%29%5E%7B-4%7D&id=WWfkM) ;
%20%3D%202#card=math&code=g%27%28x%29%20%3D%202&id=obpt9).
Conclusion: 
%20%3D(2x%2B3)%5E%7B-3%7D%20%2B%20x%20(-6)(2x%2B3)%5E%7B-4%7D%20%3D%20%5Cfrac%7B-4x%2B3%7D%7B(2x%2B3)%5E4%7D#card=math&code=h%27%28x%29%20%3D%282x%2B3%29%5E%7B-3%7D%20%2B%20x%20%28-6%29%282x%2B3%29%5E%7B-4%7D%20%3D%20%5Cfrac%7B-4x%2B3%7D%7B%282x%2B3%29%5E4%7D&id=CtewY).
Now we turn to computation of indefinite integrals. Start with an
Example: Find the anti-derivative (indefinite integral) of 
%5E3%20(2x)#card=math&code=%28x%5E2-3%29%5E3%20%282x%29&id=BletD).
Solution: 
%5E3%20(2x)%20dx#card=math&code=%5Cint%20%28x%5E2-3%29%5E3%20%282x%29%20dx&id=PTFgX) , to compute it, we can use Newton’s binomial expansion for 
,
where 
!%7D%20%3D%20%5Cbinom%7Bn%7D%7Bn-k%7D#card=math&code=%5Cbinom%7Bn%7D%7Bk%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bk%21%20%28n-k%29%21%7D%20%3D%20%5Cbinom%7Bn%7D%7Bn-k%7D&id=S2ANY) with 
integers, and 
.
%20%2B%203x%5E2%20(-3)%5E2%20%2B%20(-3)%5E3)(2x)%20%3D%202x%5E7-18x%5E4%2B54x%5E2-54x#card=math&code=%28x%5E6%2B3x%5E4%28-3%29%20%2B%203x%5E2%20%28-3%29%5E2%20%2B%20%28-3%29%5E3%29%282x%29%20%3D%202x%5E7-18x%5E4%2B54x%5E2-54x&id=DxlCg) , and
%20dx#card=math&code=%5Cint%20%282x%5E7-18x%5E4%2B54x%5E2-54x%29%20dx&id=v6XR6) = 
x%5E8%20-%20(18%2F5)x%5E5%20%2B18%20x%5E3%20-%2027%20x%5E2%20%2BC#card=math&code=%281%2F8%29x%5E8%20-%20%2818%2F5%29x%5E5%20%2B18%20x%5E3%20-%2027%20x%5E2%20%2BC&id=qhhK5)
We now apply the reverse process of the “chain rule”, namely, substitution, to compute this indefinite integral.
The derviative of 
is 
#card=math&code=%282x%29&id=ntAHz) : 
%20%3D%20(2x)%20dx#card=math&code=d%28x%5E2-%203%29%20%3D%20%282x%29%20dx&id=Qv7cs)
%5E3%20(2x)%20dx%20%3D%20%5Cint%20(x%5E2-3)%5E3%20d(x%5E2)%20%3D%20%5Cint%20(x%5E2-3)%5E3%20d(x%5E2-3)#card=math&code=%5Cint%20%28x%5E2-3%29%5E3%20%282x%29%20dx%20%3D%20%5Cint%20%28x%5E2-3%29%5E3%20d%28x%5E2%29%20%3D%20%5Cint%20%28x%5E2-3%29%5E3%20d%28x%5E2-3%29&id=KLFpL) Now we let 
%5E3%20(2x)%20dx%20%3D%20%5Cint%20(x%5E2-3)%5E3%20d(x%5E2-3)%20%3D%20%5Cint%20t%5E3%20dt%20%3D%20t%5E4%2F4%2BC%20%3D%20(1%2F4)(x%5E2-3)%5E4%20%2BC#card=math&code=%5Cint%20%28x%5E2-3%29%5E3%20%282x%29%20dx%20%3D%20%5Cint%20%28x%5E2-3%29%5E3%20d%28x%5E2-3%29%20%3D%20%5Cint%20t%5E3%20dt%20%3D%20t%5E4%2F4%2BC%20%3D%20%281%2F4%29%28x%5E2-3%29%5E4%20%2BC&id=tH6Fx)
Remark: if one uses 
#card=math&code=2xdx%3Dd%28x%5E2%29&id=NZeof) , then we let 
: 
%5E3%20du#card=math&code=%5Cint%20%28u-3%29%5E3%20du&id=bcs8c). Now either using 
#card=math&code=du%20%3D%20d%28u-3%29&id=sDklC) or expanding 
%5E3#card=math&code=%28u-3%29%5E3&id=elX9S)
Example 5.30
(6x)dx = 3 (2x)dx = 3d(x^2) . If we let 
, then 
%5E4%20%3D%20(3t%2B4)%5E4#card=math&code=%283x%5E2%2B4%29%5E4%20%3D%20%283t%2B4%29%5E4&id=nQZmc) we still have “3dt”
%20%3Dd%20(3t%2B4)#card=math&code=3dt%20%3D%20d%283t%29%20%3Dd%20%283t%2B4%29&id=gs0Hy)
Now we let 
, then 
, and
%2BC%20%3D%20(1%2F5)(3x%5E2%2B4)%5E5%2BC#card=math&code=%5Cint%20y%5E4%20dy%20%3D%20%28y%5E5%2F5%29%2BC%20%3D%20%281%2F5%29%283x%5E2%2B4%29%5E5%2BC&id=T8Qbz)
Exercise 5.25 (below Example 5.30)
%5E2%20(3x%5E2)dx#card=math&code=%5Cint%20%28x%5E3-3%29%5E2%20%283x%5E2%29dx&id=asoC8) . The key is to recognise that 
%20%3D%20d(x%5E3-3)#card=math&code=3x%5E2%20dx%20%3D%20d%28x%5E3%29%20%3D%20d%28x%5E3-3%29&id=Na7eb) . Then we let 
.
we get 
t%5E3%20%2B%20C%20%3D%20(1%2F3)(x%5E3-3)%5E3%20%2BC#card=math&code=%5Cint%20t%5E2%20dt%20%3D%20%281%2F3%29t%5E3%20%2B%20C%20%3D%20%281%2F3%29%28x%5E3-3%29%5E3%20%2BC&id=IAE07)
Example 5.31 (one has to adjust the constant factor here)
%5Cint%20(z%5E2-5)%5E%7B1%2F2%7D%202zdz#card=math&code=%281%2F2%29%5Cint%20%28z%5E2-5%29%5E%7B1%2F2%7D%202zdz&id=W7fPS) and we see that 
#card=math&code=2zdz%20%3D%20d%28z%5E2-5%29&id=omixq) . So the original integral is
%20%5Cint%20(z%5E2-5)%5E%7B1%2F2%7D%20d%20(z%5E2-5)%20%3D%20(1%2F2)%20%2F(3%2F2)%20(z%5E2-5)%5E%7B3%2F2%7D%2BC%20%3D%20(1%2F3)(z%5E2-5)%5E%7B3%2F2%7D%2BC#card=math&code=%281%2F2%29%20%5Cint%20%28z%5E2-5%29%5E%7B1%2F2%7D%20d%20%28z%5E2-5%29%20%3D%20%281%2F2%29%20%2F%283%2F2%29%20%28z%5E2-5%29%5E%7B3%2F2%7D%2BC%20%3D%20%281%2F3%29%28z%5E2-5%29%5E%7B3%2F2%7D%2BC&id=rsXRB)
In the coming week we continue to learn how to compute indefinite integrals. We shall learn two major techniques: one is integration by parts, the other is the intergration of rational functions.
(1) We have learned some basic integrals: integrals of basic elementary funcitons ( five types, powers, exp, logarithm, trigonometric, and inverse of trigonometric functions) .
In what follows we will learn the below methods
(2) The technique of substitution.
(3) integration by parts and
(4) intergration of rational functions.
