Post on 25-Sep-2020
transcript
U-substitutions
November 24, 2015 1 / 10
An example:
Computed
dx[sin(x2)]
= cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
cos(x2)2xdx = sin(x2)
+ C .
What’s missing?
November 24, 2015 2 / 10
An example:
Computed
dx[sin(x2)] = cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
cos(x2)2xdx = sin(x2)
+ C .
What’s missing?
November 24, 2015 2 / 10
An example:
Computed
dx[sin(x2)] = cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
cos(x2)2xdx = sin(x2)
+ C .
What’s missing?
November 24, 2015 2 / 10
An example:
Computed
dx[sin(x2)] = cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative.
∫cos(x2)2xdx = sin(x2)
+ C .
What’s missing?
November 24, 2015 2 / 10
An example:
Computed
dx[sin(x2)] = cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
cos(x2)2xdx =
sin(x2)
+ C .
What’s missing?
November 24, 2015 2 / 10
An example:
Computed
dx[sin(x2)] = cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
cos(x2)2xdx = sin(x2)
+ C .
What’s missing?
November 24, 2015 2 / 10
An example:
Computed
dx[sin(x2)] = cos(x2) · 2x
Compute
∫cos(x2)2xdx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
cos(x2)2xdx = sin(x2) + C .
What’s missing?
November 24, 2015 2 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))]
= . . .The chain rule · · · = f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
f ′(g(x)) · g ′(x)dx = f (g(x)) + C .
November 24, 2015 3 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))] = . . .The chain rule · · · =
f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
f ′(g(x)) · g ′(x)dx = f (g(x)) + C .
November 24, 2015 3 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))] = . . .The chain rule · · · = f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
f ′(g(x)) · g ′(x)dx = f (g(x)) + C .
November 24, 2015 3 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))] = . . .The chain rule · · · = f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
f ′(g(x)) · g ′(x)dx = f (g(x)) + C .
November 24, 2015 3 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))] = . . .The chain rule · · · = f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative.
∫f ′(g(x)) · g ′(x)dx = f (g(x)) + C .
November 24, 2015 3 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))] = . . .The chain rule · · · = f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
f ′(g(x)) · g ′(x)dx =
f (g(x)) + C .
November 24, 2015 3 / 10
The U-substitution rule: The chain rule reversed.
Let f and g be differentiable functions. Compute
d
dx[f (g(x))] = . . .The chain rule · · · = f ′(g(x))g ′(x)
Compute
∫f ′(g(x)) · g ′(x)dx .
Notice that we are very lucky! The derivative computation above gives usan antiderivative. ∫
f ′(g(x)) · g ′(x)dx = f (g(x)) + C .
November 24, 2015 3 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du =
g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du =
F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C =
F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Why U-substitution? What does it meanSuppose that we have an antiderivative F (x) for f (x).
We will again compute
∫f (g(x)) · g ′(x)dx
Let’s substitute u = g(x) so that du = g ′(x)dx .
Making these substitutions:
∫f (g(x)) · g ′(x)dx =
∫f (u) · du
We have an antiderivative for f .
∫f (u) · du = F (u) + C = F (g(x)) + C .
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION:
1 After you substitute in u = g(x), x must no longer appears in theintegrand!
2 Make sure you back substitute! The indefinite integral of f (x) is afunction in x , not a function in u.
November 24, 2015 4 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.
So du = 2x · dx . Make these substitutions:∫(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du =
2x · dx . Make these substitutions:∫(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx .
Make these substitutions:∫(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u!
Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Example:
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
Compute
∫(x2 − 1)32xdx .
Substitute u = x2 − 1.So du = 2x · dx . Make these substitutions:∫
(x2 − 1)32xdx =
∫u3du =
u4
4+ C .
We aren’t done! We want a function in x , not a function in u! Backsubstitute u = x2 − 1:∫
(x2 − 1)32xdx =u4
4+ C =
(x2 − 1)4
4+ C .
Double Check: Compute a derivative.
November 24, 2015 5 / 10
Some guided examples
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
CAUTION: After you substitute in u, x no longer appears in theintegrand!!!!!!
1 Compute
∫1
x4 + 1· 4x3dx (Use u = x4 + 1)
2 Compute
∫1
x4 + 1· 2xdx (Use u = x2)
3 Compute
∫ √5x − 2 · 5dx (Use u = 5x − 1)
4 Compute
∫e5xdx (Use u = 5x . You will need to be creative to find
du in the integrand.)
November 24, 2015 6 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x).
Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du =
− sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx .
Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?
Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du =
sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dx
Make this substitution:∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu =
− ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C =
ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!
November 24, 2015 7 / 10
Sometimes u is harder to pick
Compute
∫tan(x)dx =
∫sin(x)
cos(x)dx .
Here’s a trick:
∫sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx .
Let’s try u = cos(x). Then du = − sin(x)dx . Do you see −sin(x)dx in theintegrand?Not quite. Instead we see −du = sin(x)dxMake this substitution:∫
sin(x)
cos(x)dx =
∫1
cos(x)· sin(x)dx =
∫−1
udu = − ln(|u|) + C
We want an answer in terms of x , not in terms of u. Back substituteu = cos(x). ∫
sin(x)
cos(x)dx = − ln(|u|) + C = ln(| cos(x)|) + C
Don’t forget to take a derivative to double check!November 24, 2015 7 / 10
Less guided examples
Theorem
if u = g(x) is a differentiable function then
∫f (g(x))g ′(x)dx =
∫f (u)du
1 Compute
∫e−3x+2dx
2 Compute
∫sin(x) · cos(x)2dx
3 Compute
∫(x − 5)1,000dx .
November 24, 2015 8 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root.
Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .
Rewrite the integral so as to see du.∫ √x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.
∫ √x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?
Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting.
Compute
∫ √x2 + 2 · x3dx
The most complicated stuff is inside the square root. Set u = x2 + 2.Thus, du = 2xdx .Rewrite the integral so as to see du.∫ √
x2 + 2 · x3dx =
∫ √x2 + 2 · x2 1
22xdx
We have an extrax2
2sitting around. We need to be able to express this in
terms of u’s. Can we?Recall u = x2 + 2 so that u − 2 = x2
Now we can substitute away all the x ’s:∫ √x2 + 2 · x3dx =
∫ √x2 + 2
x2
22xdx =
∫ √u
(u − 2)
2du
More on the next slide . . . .
November 24, 2015 9 / 10
Sometimes There are parts left over after you seem to bedone substituting. Part II.
Last slide: After there substitution u = x2 + 2 we got∫ √x2 + 2 · x3dx =
∫ √u
(u − 2)
2du
=
∫1
2u3/2 − u1/2du
=
(1
2· 2
5u5/2 − 2
3u3/2
)+ C
=1
5u5/2 − 2
3u3/2 + C
Back substitute∫ √x2 + 2x7dx =
1
5(x2 + 2)5/2 − 2
3(x2 + 2)3/2 + C
We should double check our answer.
November 24, 2015 10 / 10
Sometimes There are parts left over after you seem to bedone substituting. Part II.
Last slide: After there substitution u = x2 + 2 we got∫ √x2 + 2 · x3dx =
∫ √u
(u − 2)
2du
=
∫1
2u3/2 − u1/2du
=
(1
2· 2
5u5/2 − 2
3u3/2
)+ C
=1
5u5/2 − 2
3u3/2 + C
Back substitute∫ √x2 + 2x7dx =
1
5(x2 + 2)5/2 − 2
3(x2 + 2)3/2 + C
We should double check our answer.
November 24, 2015 10 / 10
Sometimes There are parts left over after you seem to bedone substituting. Part II.
Last slide: After there substitution u = x2 + 2 we got∫ √x2 + 2 · x3dx =
∫ √u
(u − 2)
2du
=
∫1
2u3/2 − u1/2du
=
(1
2· 2
5u5/2 − 2
3u3/2
)+ C
=
1
5u5/2 − 2
3u3/2 + C
Back substitute∫ √x2 + 2x7dx =
1
5(x2 + 2)5/2 − 2
3(x2 + 2)3/2 + C
We should double check our answer.
November 24, 2015 10 / 10
Sometimes There are parts left over after you seem to bedone substituting. Part II.
Last slide: After there substitution u = x2 + 2 we got∫ √x2 + 2 · x3dx =
∫ √u
(u − 2)
2du
=
∫1
2u3/2 − u1/2du
=
(1
2· 2
5u5/2 − 2
3u3/2
)+ C
=1
5u5/2 − 2
3u3/2 + C
Back substitute∫ √x2 + 2x7dx =
1
5(x2 + 2)5/2 − 2
3(x2 + 2)3/2 + C
We should double check our answer.
November 24, 2015 10 / 10
Sometimes There are parts left over after you seem to bedone substituting. Part II.
Last slide: After there substitution u = x2 + 2 we got∫ √x2 + 2 · x3dx =
∫ √u
(u − 2)
2du
=
∫1
2u3/2 − u1/2du
=
(1
2· 2
5u5/2 − 2
3u3/2
)+ C
=1
5u5/2 − 2
3u3/2 + C
Back substitute∫ √x2 + 2x7dx =
1
5(x2 + 2)5/2 − 2
3(x2 + 2)3/2 + C
We should double check our answer.
November 24, 2015 10 / 10
Sometimes There are parts left over after you seem to bedone substituting. Part II.
Last slide: After there substitution u = x2 + 2 we got∫ √x2 + 2 · x3dx =
∫ √u
(u − 2)
2du
=
∫1
2u3/2 − u1/2du
=
(1
2· 2
5u5/2 − 2
3u3/2
)+ C
=1
5u5/2 − 2
3u3/2 + C
Back substitute∫ √x2 + 2x7dx =
1
5(x2 + 2)5/2 − 2
3(x2 + 2)3/2 + C
We should double check our answer.
November 24, 2015 10 / 10