U-substitution
→ Chain Rule
Published in
3 min readJan 21, 2019
The
u-substitution
is to solve an integral of composite function, which is actually to UNDO theChain Rule
.
▶ Back to previous note on: Chain Rule
Compare how we handle the composite functions with derivatives & integrals:
- For taking the derivative of a COMPOSITE function, we apply the
Chain rule
. - For taking the integral of a COMPOSITE function, we apply the
u-substitution
.
Refer to Khan academy: 𝘶-substitution: defining 𝘶
We use u-substitution
when we need to integrate an expression of the form of:
Strategy:
- Find a function as
u
- Find or MAKE an
u'
at the outside so that you can pairu'
withdx
- Replace
u' · dx
withdu
, becauseu' = du/dx
- Rewrite the Integral in term of
u
, and calculate the integral - Back substitute the function of
u
back to the result.
How to select u
Selecting u
is the most tricky part here.
Example
Solve:
- Apparently, we ignore the wrapper
sin()
here. - We notice that the derivative of
-x+2
is-1
which we could find it at outside. - So let
u = -x+2
andu' = -1
- So rewrite the integral to
ʃ sin(u) · u' · dx = ʃ sin(u) · du
- It looks quite neat, so the
u = -x+2
is alright.
Example
Solve:
- Apparently it’s in form of
ʃ u'/u · dx
- So that we can make
u'·dx = du
and the integral becomesʃ 1/u · du
- Quite nice, so the answer would be out of there.
How to calculate Indefinite Integral with u-substitution
Example
Solve:
- With a real quick eyeballing, we see it’s in form of
ʃ u' · u⁶ · dx
- So with
u' · dx = du
we will get the simplified formʃ u⁶ · du = u⁷/7
- Back substitute function of u back to get the result:
How to calculate Definite Integral with u-substitution
Example
Solve:
Example (self-made u’)
Solve:
Example (Inverse Trig Rule)
Solve:
- Notice this radical form should directly use the Reversed Inverse Trig Rule:
- So that we assume
a = 1 & u = 3x
. - Since
u' = 3
so we need to make a3
from nowhere. - Rewrite the formula to:
1/3 ʃ 3/(1+u²) ·dx = 1/3 ʃ 1/(1+u²) ·du
- Apply the Reversed Inverse Trig Rule to get:
1/3 arctan(u) + C
- Back substitute
3x
tou
and the boundaries back tox
get the resultπ/6
.