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Chapter 20
Substitution
20.1
Integration by guessing
20.1.1
Idea
Sometimes we have a notion of the form of an antiderivative. That allows
us to take a guess and then work out the details and adjust to get the right
answer.
20.1.2
Example
What is
Z
sin(2x) dx
The rules that we have worked out so far don’t help. But the derivative of
cos is -sin so we can try
Z
sin(2x) dx = − cos(2x) + C
Taking the derivative of the right-hand side we get 2 sin(2x) which is not
exactly what we want but if we now try
Z
1
sin(2x) dx = − cos(2x) + C
2
this works.
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20.1.3
CHAPTER 20. SUBSTITUTION
Example
Z
x sin(x2 ) dx
We might reason like this: The derivative of cos(x2 ) is −2x sin(x2 ) so we
could try
Z
x sin(x2 ) dx = cos(x2 )
Taking the derivative of the suggested answer gives us something that is off
by a constant factor. What is the correct answer?
20.2
Differentials
20.2.1
Idea
Symbols like dx and dy don’t represent real numbers and dy/dx is not really
a fraction. But in some cases we can formally manipulate these symbols as
if the were numbers.
20.2.2
Differentials
If
df
= f′
dx
then the differential df is defined to be
df = f ′ dx
20.2.3
Example
If u = sin(x2 ) then du = 2x cos(x2 )dx.
If u = x2 + x + 1 then du = (2x + 1)dx.
20.3
Substitution
20.3.1
Chain rule
Recall the chain rule:
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20.3. SUBSTITUTION
d
f (g(x)) = f ′ (g(x))g ′ (x)
dx
It follows that
Z
20.3.2
f ′ (g(x))g ′ (x) dx = f (g(x)) + C
Idea
Integrals whose arguments look like something given by the chain rule can
be made simpler using substitution and differentials.
20.3.3
Example
I =
Z
x(x2 + 7)12 dx
We see that the derivative of the term in round brackets is outside of the
brackets (up to a multiplicative constant). This suggests that the integral
might be done keeping the chain rule in mind and using substitutions and
differentials to simplify the problem.
So, let
u = x2 + 7
and then
du = 2xdx
and
xdx =
1
du
2
Then I may be rewritten
I =
Z
u12 /2 du
and the solution of that is
I = u13 /26 + C
But the original problem was in terms of x so the final answer so also be
expressed in terms of x so we substitute for u to get
I = (x2 + 7)13 /26 + C
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20.3.4
CHAPTER 20. SUBSTITUTION
Example
I =
Z
sin(3x) cos(3x) dx
We can try
u = sin(3x)
du = 3 cos(3x)dx
cos(3x)dx =
1
du
3
Complete the problem.
20.3.5
Example
I =
Z
√
x x + 1 dx
We can try
u = x+1
du = dx
xdx = (u − 1)du
Complete the problem.