6.9 - Integrating Using Substitution

Connection to the Chain Rule

Chain rule review: for composite functions, the derivative is

\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)

That’s the derivative of the outside times the derivative of the inside.

U-substitution is the reverse process, for integrals involving composite functions.

When to Use U-Substitution

Rule of thumb: if taking the derivative would require the chain rule, then taking the integral requires u-substitution.

The integrand is the expression being integrated, meaning everything between the

\int

and the

dx

\int \underbrace{f(x)}_{\text{integrand}} \, dx

The U-Substitution Method

Basic Procedure

Choose $u$ : let $u$ be the inside function, usually the inner part of a composite

Find $du$ : take the derivative, $\frac{du}{dx} =$ derivative of $u$ , then solve for $dx$ : $dx = \frac{du}{\text{derivative}}$

Substitute: replace all the $x$ terms with $u$ and $dx$ with the expression from step 2

Integrate: integrate with respect to $u$

Back-substitute: replace $u$ with the original expression in terms of $x$

Note: the letter

u

is just convention. Any letter works.

Example 1: Basic U-Substitution

Evaluate

\int (3x-4)^5 \, dx

Step 1: Choose

u

u = 3x - 4

Step 2: Find

du

and solve for

dx

\frac{du}{dx} = 3 \quad \Rightarrow \quad du = 3 \, dx \quad \Rightarrow \quad dx = \frac{du}{3}

Step 3: Substitute.

\int (3x-4)^5 \, dx = \int u^5 \cdot \frac{du}{3}

Step 4: Integrate.

= \frac{1}{3} \int u^5 \, du = \frac{1}{3} \cdot \frac{u^6}{6} + C = \frac{u^6}{18} + C

Step 5: Back-substitute.

= \frac{(3x-4)^6}{18} + C

Example 2: When Terms Cancel

Evaluate

\int 6x^2(x^3 + 4)^5 \, dx

Step 1: Choose

u

u = x^3 + 4

Why this choice? The derivative will be

3x^2

, which will cancel with the

6x^2

already sitting in the integrand.

Step 2: Find

du

and solve for

dx

\frac{du}{dx} = 3x^2 \quad \Rightarrow \quad du = 3x^2 \, dx \quad \Rightarrow \quad dx = \frac{du}{3x^2}

Step 3: Substitute.

\int 6x^2 \cdot u^5 \cdot \frac{du}{3x^2}

Step 4: Simplify, and the

x^2

terms cancel.

= \int \frac{6}{3} u^5 \, du = 2\int u^5 \, du

Step 5: Integrate.

= 2 \cdot \frac{u^6}{6} + C = \frac{u^6}{3} + C

Step 6: Back-substitute.

= \frac{(x^3+4)^6}{3} + C

Key insight: the

x

terms from the original integrand should cancel with terms from

dx

. That’s your signal that you picked the right

u

Example 3: With Square Roots

Evaluate

\int \dfrac{(\sqrt{x}-1)^2}{\sqrt{x}} \, dx

Step 1: Choose

u

u = \sqrt{x} - 1

Step 2: Find

du

\frac{du}{dx} = \frac{1}{2\sqrt{x}} \quad \Rightarrow \quad du = \frac{1}{2\sqrt{x}} \, dx \quad \Rightarrow \quad dx = 2\sqrt{x} \, du

Step 3: Substitute.

\int \frac{u^2}{\sqrt{x}} \cdot 2\sqrt{x} \, du

Step 4: Cancel, since the

\sqrt{x}

terms drop out.

= \int 2u^2 \, du

Step 5: Integrate.

= 2 \cdot \frac{u^3}{3} + C = \frac{2u^3}{3} + C

Step 6: Back-substitute.

= \frac{2(\sqrt{x}-1)^3}{3} + C

Example 4: Exponential with Trig

Evaluate

\int \sin(x) e^{\cos(x)} \, dx

Test the choices:

Bad choice:

u = \sin(x)

, which gives

du = \cos(x) \, dx

. That won’t cancel the

\sin(x)

in front.

Good choice:

u = \cos(x)

, which gives

du = -\sin(x) \, dx

. That WILL cancel with the

\sin(x)

in the integrand.

Solution:

u = \cos(x), \quad du = -\sin(x) \, dx, \quad dx = \frac{du}{-\sin(x)}

Substitute:

\int \sin(x) \cdot e^u \cdot \frac{du}{-\sin(x)} = \int -e^u \, du

= -e^u + C = -e^{\cos(x)} + C

Double U-Substitution

Sometimes you need u-substitution twice in the same problem.

Example: $\int \cot(3x) \, dx$

First substitution:

u = 3x, \quad du = 3 \, dx, \quad dx = \frac{du}{3}

\int \cot(3x) \, dx = \frac{1}{3}\int \cot(u) \, du

Problem: we don’t have a formula for

\int \cot(u) \, du

Solution: rewrite cotangent.

\int \cot(u) \, du = \int \frac{\cos(u)}{\sin(u)} \, du

Second substitution (the “w” substitution):

w = \sin(u), \quad dw = \cos(u) \, du, \quad du = \frac{dw}{\cos(u)}

\int \frac{\cos(u)}{\sin(u)} \, du = \int \frac{\cos(u)}{w} \cdot \frac{dw}{\cos(u)} = \int \frac{1}{w} \, dw

= \ln|w| + C = \ln|\sin(u)| + C = \ln|\sin(3x)| + C

Final answer:

\int \cot(3x) \, dx = \frac{1}{3}\ln|\sin(3x)| + C

Advanced Technique: Solving for $x$

Sometimes you have to solve the

u

equation for

x

Example: $\int \dfrac{x}{\sqrt{x+1}} \, dx$

Substitution:

u = x + 1, \quad du = dx

Problem: there’s still an

x

in the numerator after substituting.

Solution: solve for

x

in terms of

u

u = x + 1 \implies x = u - 1

Substitute everything:

\int \frac{x}{\sqrt{x+1}} \, dx = \int \frac{u-1}{\sqrt{u}} \, du

Separate the fraction:

= \int \left(\frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}}\right) du = \int \left(u^{1/2} - u^{-1/2}\right) du

Integrate:

= \frac{u^{3/2}}{3/2} - \frac{u^{1/2}}{1/2} + C = \frac{2u^{3/2}}{3} - 2u^{1/2} + C

Back-substitute:

= \frac{2(x+1)^{3/2}}{3} - 2(x+1)^{1/2} + C

Recognizing Inverse Trig (Not U-Substitution)

Problem:

\int \dfrac{1}{\sqrt{1-4x^2}} \, dx

Common mistake: trying

u = 1 - 4x^2

, which gives

du = -8x \, dx

. But there’s no

x

in the numerator to cancel with, so this goes nowhere.

Correct approach: recognize the inverse trig pattern. Rewrite as

\frac{1}{\sqrt{1-(2x)^2}}

This matches the pattern

\frac{1}{\sqrt{1-u^2}}

, which integrates to

\sin^{-1}(u)

Solution:

u = 2x, \quad du = 2 \, dx, \quad dx = \frac{du}{2}

\int \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{2} = \frac{1}{2}\sin^{-1}(u) + C = \frac{1}{2}\sin^{-1}(2x) + C

U-Substitution with Definite Integrals

For definite integrals, change the limits of integration instead of back-substituting.

Procedure

Do the u-substitution as usual

Convert the limits: plug the original limits into $u = g(x)$ to get the new limits

Integrate with respect to $u$ using the new limits

Evaluate, with no need to substitute back

Example: $\int_0^2 t^2\sqrt{t^3+1} \, dt$

Step 1: Substitution.

u = t^3 + 1, \quad du = 3t^2 \, dt, \quad dt = \frac{du}{3t^2}

Step 2: Change the limits.

When

t = 0

u = 0^3 + 1 = 1

When

t = 2

u = 2^3 + 1 = 9

New limits: from 1 to 9.

Step 3: Substitute and simplify.

\int_0^2 t^2 \cdot \sqrt{u} \cdot \frac{du}{3t^2} = \int_1^9 \frac{1}{3}u^{1/2} \, du

Step 4: Integrate.

= \frac{1}{3} \cdot \frac{u^{3/2}}{3/2} \bigg|_1^9 = \frac{2}{9} u^{3/2} \bigg|_1^9

Step 5: Evaluate using the new limits.

= \frac{2}{9}\left[9^{3/2} - 1^{3/2}\right] = \frac{2}{9}[27 - 1] = \frac{2}{9}(26) = \frac{52}{9}

Key advantage: no need to convert back to

t

, because you changed the limits.

Example 2: $\int_0^{\pi/2} \sin^3(x)\cos(x) \, dx$

Substitution:

u = \sin(x), \quad du = \cos(x) \, dx

Change the limits:

When

x = 0

u = \sin(0) = 0

When

x = \frac{\pi}{2}

u = \sin\left(\frac{\pi}{2}\right) = 1

New limits: from 0 to 1.

Integrate:

\int_0^1 u^3 \, du = \frac{u^4}{4}\bigg|_0^1 = \frac{1}{4} - 0 = \frac{1}{4}

Key Takeaways

U-substitution reverses the chain rule for integration

When to use it: if the derivative would need the chain rule, the integral needs u-sub

Choose $u$ to be the inside function of a composite

Check your choice: the derivative of $u$ should cancel with existing terms

Always solve for $dx$ : $dx = \frac{du}{\text{derivative of } u}$

Terms should cancel. After substituting, all the $x$ terms should be gone

Back-substitute to replace $u$ with the original expression for indefinite integrals

For definite integrals, change the limits instead of back-substituting

Double substitution is sometimes needed for harder problems (the w-substitution)

Solve for $x$ when needed, using $u = g(x)$ to write $x$ in terms of $u$

Not everything is u-sub. Some patterns are inverse trig formulas, so learn the difference

Think ahead: before choosing $u$ , check whether its derivative will actually help you cancel

Back Next

6.9 - Integrating Using Substitution

Connection to the Chain Rule

When to Use U-Substitution

The U-Substitution Method

Basic Procedure

Example 1: Basic U-Substitution

Example 2: When Terms Cancel

Example 3: With Square Roots

Example 4: Exponential with Trig

Double U-Substitution

Example: ∫cot⁡(3x) dx\int \cot(3x) \, dx∫cot(3x)dx

Advanced Technique: Solving for xxx

Example: ∫xx+1 dx\int \dfrac{x}{\sqrt{x+1}} \, dx∫x+1​x​dx

Recognizing Inverse Trig (Not U-Substitution)

U-Substitution with Definite Integrals

Procedure

Example: ∫02t2t3+1 dt\int_0^2 t^2\sqrt{t^3+1} \, dt∫02​t2t3+1​dt

Example 2: ∫0π/2sin⁡3(x)cos⁡(x) dx\int_0^{\pi/2} \sin^3(x)\cos(x) \, dx∫0π/2​sin3(x)cos(x)dx

Key Takeaways

Example: $\int \cot(3x) \, dx$

Advanced Technique: Solving for $x$

Example: $\int \dfrac{x}{\sqrt{x+1}} \, dx$

Example: $\int_0^2 t^2\sqrt{t^3+1} \, dt$

Example 2: $\int_0^{\pi/2} \sin^3(x)\cos(x) \, dx$