U-substitution → Chain Rule
Last updated
Last updated
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 pair u'
with dx
Replace u' · dx
with du
, because u' = du/dx
Rewrite the Integral in term of u
, and calculate the integral
Back substitute the function of u
back to the result.
Selecting u
is the most tricky part here.
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
and u' = -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.
Apparently it's in form of ʃ u'/u · dx
So that we can make u'·dx = du
and the integral becomes ʃ 1/u · du
Quite nice, so the answer would be out of there.
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 form ʃ u⁶ · du = u⁷/7
Back substitute function of u back to get the result:
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 a 3
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
to u
and the boundaries back to x
get the result π/6
.
Solve:
Solve:
Solve:
Solve:
Solve:
Solve: