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