Integration using Substitution

Section 4.6 Integration using Substitution

Standards

Integrate using substitution

We learned that we can use the scalar multiple and sum properties of derivatives to calculate anti-derivatives. However, neither the product rule of derivatives nor anything else produces a product rule for integrals. Now we will consider how we can apply the chain rule of derivatives to find more anti-derivatives.

Subsection 4.6.1 Using the Chain Rule backwards

Activity 4.6.1. Reversing the Chain Rule.

The goal of this activity is to recognize how we can use the Chain Rule to find anti-derivatives.

(a)

What is the derivative of \(\sin u\text{?}\)

(b)

What is \(\int \cos u \; du \text{?}\)

(c)

What is the derivative of \(\sin (x^2)\text{?}\)

(d)

What is \(\int \cos(x^2)2x \; dx \)

Activity 4.6.2. Identifying Substitutions.

The goal of this activity is to learn to identify substitutions for finding anti-derivatives.

(a)

For each of the following integrals determine first if a substitution could be used. Second identify the substitution \(u\text{.}\)

\(\displaystyle \int \cos(x^2+5x+3)(2x+5) \; dx\)
\(\displaystyle \int e^{\sin x} \cos x \; dx\)
\(\displaystyle \int \sin^{76} \theta \cos\theta \; d\theta\)
\(\displaystyle \int (x^3-x)^{14}3x^2 \; dx\)

Subsection 4.6.2 Examples of integral substitution

Example 4.6.1. Using substitution.

\(\int 2e^{2x+1} \; dx\)

Solution.

\begin{align*} \int 2e^{2x+1} \; dx \amp = \amp \begin{array}{rcl} u \amp = \amp 2x+1 \\ du \amp = \amp 2 \; dx \end{array}\\ \int e^u \; du \amp = \amp e^u+C\\ \amp = \amp e^{2x+1}+C. \end{align*}

Example 4.6.2. Substitutions for scalars.

\(\int e^{5x} \; dx\)

Solution.

\begin{align*} \int e^{5x} \; dx \amp = \amp \begin{array}{rcl} u \amp = \amp 5x \\ du \amp = \amp 5 \; dx \end{array}\\ \frac{1}{5}\int e^{5x} 5 \; dx \amp =\\ \frac{1}{5}\int e^u \; du \amp = \amp \frac{1}{5}e^u+C\\ \amp = \amp \frac{1}{5}e^{5x}+C. \end{align*}

Example 4.6.3. Substitutions for Definite Integrals.

\(\int_1^{\sqrt{2}} \sin(\pi x^2)2\pi x \; dx\)

Solution.

\begin{align*} \int_0^{1/\sqrt{2}} \sin(\pi x^2)2\pi x \; dx \amp = \amp \begin{array}{rcl} u \amp = \amp \pi x^2 \\ du \amp = \amp 2\pi x \; dx \end{array}\\ \int_0^{\pi/2} \sin(u) \; du \amp = \amp \left. -\cos(u) \right|_0^{\pi/2} \\ \amp = \amp -\cos(\pi/2) - [-\cos(0)]\\ \amp = \amp 1 \end{align*}

Example 4.6.4. Multipart Substitutions.

\(\int (x-1)\sqrt{x+1} \; dx\)

Solution.

\begin{align*} \int (x-1)\sqrt{x+1} \; dx \amp = \amp \begin{array}{rcl} u \amp = \amp x+1 \\ du \amp = \amp dx \\ x \amp = \amp u-1 \end{array}\\ \int (u-1-1)\sqrt{u} \; du \amp =\\ \int (u-2)\sqrt{u} \; du \amp = \\ \int u^{3/2}-2u^{1/2} \; du \amp = \amp \frac{2}{5}u^{5/2}-\frac{4}{3}u^{3/2}+C\\ \amp = \amp \frac{2}{5}(x+1)^{5/2}-\frac{4}{3}(x+1)^{3/2}+C. \end{align*}

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