Sep 16th : Substitutions for integrals containing the expression
√
ax2 + bx + c
Abstract:
• integral of some standard forms:
Z p
1 − u2 du;
Z p
u2 − 1du;
Z p
u2 + 1du
For simplicity, we will assume u ≥ 0 in the following discussion.
√
• using substitution to reformulate the general form: ax2 + bx + c into standard forms.
standard case 1:
R√
1 − u2 du
The annoying part of this integral is the ”square term” inside the square root. Somehow, you want
to reformulate the inside term into a perfect square, so that the square root would be cancelled.
And you has also observed that 1 − u2 should ≥ 0, then |u| ≤ 1. Wait, what does this motivate
you? Trignometric functions, sin(θ), cos(θ)! Following our intuition, we would do the substitution:
u = cos(θ), (−π/2 ≤ θ < π/2). Then we have:
Z p
Z
2
1 − u du = sin(θ)(−sin(θ))dθ
Z
cos(2θ) − 1
=
dθ
2
sin(2θ) θ
=
− +C
4
2
√
2
u 1−u
arccos(u)
=
−
+C
2
2
standard case 2:
R√
u2 − 1du
Similary with the first case, we have noticed the annoying part is still the ”square term” inside
the square root. We still want to reformulate the inside term into a perfect square. But wait, see
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the slight difference from case 1? Here we have: u2 − 1 ≥ 0, so |u| ≥ 1. So direct substitution of
trignometric function: sin(θ), cos(θ) doesn’t work here! But have you noticed the symmetry? If
|u| ≥ 1, then how about
1
u?
Yes, that’s the way we do it. We can do the substitution: u =
1
cos(θ) .
Then
Z p
Z s
sin2 (θ)
u2 − 1du = −
sin(θ)dθ
cos2 (θ)
Z
−sin2 (θ)
=
dx
cos(θ)
Z
1
dx
= cos(θ) −
cos(θ)
sin(2θ) θ
= −sin(θ) −
+ +C
4√
2
p
2
1
−
u
u
arccos(u)
= − 1 − u2 −
+
+C
2
2
We can solve case 2 perfectly by the method we used above. But here is another trick to solve
case 2: using rational substitution.
• rational substitution:
Still we assume u ≥ 0. Then we can do the substitution: u = 12 (t + 1t ), t > 0. Then we have:
Z r
Z p
1
1
1 2
2
u − 1du =
(t + 2 − 2)(1 − 2 )dt
4
t
t
Z
1
1
1
=
(t − )(1 − 2 )dt
2
t
t
Z
1
2
1
=
t − + 3 dt
2
t
t
2
t
1
=
− ln(t) − 2 + C
4
4t
standard case 3:
R√
u2 + 1du
Now, after you have finished two standard cases, you have grown into an experience man. Any
ideas for case 3? (What is the annoying part? How to cancel it?)
2
Think for a minute.
Proceeding . . .
Now, as you might expect, we can do the substitution: u = tan(θ). Then:
Z p
Z s
u2 + 1du =
1
sec2 (θ)dθ
cos2 (θ)
Z
1
dθ
cos3 (θ)
Z
1
dv
1 − v4
=
(v = sin(θ))
=
(using partial fraction now)
genearl form:
√
ax2 + bx + c
Now, let’s crack the general form. We want to transform the general quadratic polynomial into
standard form:
ax2 + bx + c = a(x +
b 2
2a )
+c−
b2
4a .
Then the next step depends on a,
b2
4a .
If a < 0,
b2
4a
> 0, then we have case 1 to solve.
If a > 0,
b2
4a
< 0, then we have case 2 to solve.
If a > 0,
b2
4a
> 0, then we have case 3 to solve.
(Think about it: what if
b2
4a
= 0? Could both a < 0 and
3
b2
4a
< 0?)
addition:
There are several facts you should be familiar with:
• scaling:
Z
(use substitution v =
√
√
1
du (a > 0)
au2 − 1
au)
1
=√
a
Z p
v 2 − 1dv
By Scaling, it has been reformuated into the standard case 2.
Z
(use substitution v =
√
1
u2
−a
du (a > 0)
√u )
a
Z p
=
v 2 − 1dv
By similar Scaling, it has been reformuated into the standard case 2.
Exercise: Think about how to do the following simpler integral by scaling them first:
Z
√
ax + bdx
Z p
2x2 + 1dx
• basic functions:
4
Z r
3x2
1
dx
+ 6x + 6
There are roughly four types of basic functions that could describe our world to a great extent:
• polynomial and rational function:
e.g, an xn + an−1 xn−1 + . . . + a1 x + a0 ;
an xn +...+a0
bm xm +...+b0 ;
These are the most basic functions, just like numbers. They are easily calculated.
• exponential function:
e.g, ex ;
These functions describe fast proliferation, such as fission.
• logrithim:
e.g, ln(x);
The inverse function of exponential function.
• trignometric function:
e.g, sin(x), cos(x), tran(x), cot(x);
These functions are used to describe oscillation
.
You should definitely go familiar with these functions, (know their derivative, integral, etc). These
would definitely help you a lot for your calculus study.
Summary
Important things you should know:
√
This section is mainly talking about substitution tricks for integrals that contain the term: ax2 + bx + c.
√
• If the integral has a term of 1 − x2 , think about using the substitution: x = sin(θ); (or cos(θ))
√
1
1
• If the integral has a term of x2 − 1, think about using the substitution: x = cos(θ)
; (or sin(θ)
,
or t + 1t )
• If the integral has a term of
√
x2 + 1, think about using the substitution: x = tan(θ); (or t − 1t )
But those substitutions are not necessary to be the right choice. Think about the following one:
R
√ x
dx.
1−x2
u = x2 is the correct substitution, while x = sin(θ) does not help to solve this integral.
So take a look at the integral carefully before you choose your substitution!
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