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Section 4.6 Integration 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 41 . 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 42 . 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.
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.
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*}
