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Chapter 5: Q. 13 (page 428)

If $u (x) = \sin 3 x$ and $v (x) = x$ , what are du and dv? Write down $鈭� u d v$ and $鈭� v d u$ in this situation. Which of these integrals would be easier to find? What does this exercise have to do with integration by parts?

Short Answer

Expert verified

$d u = 3 \cos 3 x d x$ , $d v = d x$
$鈭� u d v = 鈭� \sin 3 x d x$ , $鈭� v d u = 鈭� 3 x \cos 3 x d x$
$鈭� u d v$ is easier to find.
$鈭� v d u$ can be solved by using integration by parts.

Step by step solution

01

Step 1. Given information

The given functions are $u (x) = \sin 3 x$ and $v (x) = x$ .

02

Step 2. Evaluate the derivatives to obtain du and dv.

Differentiate the given functions separately with respect to x.

$\begin{array}{rcl} \frac{d}{d x} u (x) & = & \cos 3 x 路 3 \\ d u & = & 3 \cos 3 x d x \end{array}$

$\begin{array}{rcl} \frac{d}{d x} v (x) & = & 1 \\ d v & = & d x \end{array}$

03

Step 3. Obtain the expression for ∫udv and ∫vdu.

Substitute the given and obtained values to obtain $鈭� u d v$ and $鈭� v d u$ .

role="math" localid="1648817334091" $鈭� u d v = 鈭� \sin 3 x d x$

$鈭� v d u = 鈭� 3 x \cos 3 x d x$

04

Step 4. Explanation

The integral $鈭� u d v = 鈭� \sin 3 x d x$ is easier to find because it involves only sine function, whereas the other integral, that is, $鈭� v d u = 鈭� 3 x \cos 3 x d x$ involves multiplication of two functions. Thus, integration by parts would be used in solving $鈭� v d u$ .

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Most popular questions from this chapter

Solve the integral $鈭� x^{3} \sqrt{x^{2} - 1} d x$ three ways:

(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;

(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$

(c) with the trigonometric substitution x = sec u.

Which of the integrals that follow would be good candidates for trigonometric substitution? If a trigonometric substitution is a good strategy, name the substitution. If another method is a better strategy, explain that method.

$(a) 鈭� \frac{4 + x^{2}}{x} d x$ $(b) 鈭� \frac{x}{4 + x^{2}} d x$

role="math" localid="1648759296940" $(c) 鈭� \frac{x^{2}}{4 + x^{2}} d x$ $(d) 鈭� \frac{16 鈭� x^{4}}{4 + x^{2}} d x$

Why is it okay to use a triangle without thinking about the unit circle when simplifying expressions that result from a trigonometric substitution with $x = a \sin u$ or $x = a \tan u$ ? Why do we need to think about the unit circle after trigonometric substitution with $x = a s e c u$ ?

Explain how to know when to use the trigonometric substitutions $x = a \sin 鈦� u, x = a \tan 鈦� u, a n d x = a \sec 鈦� u$ , Describe the trigonometric identity and the triangle that will be needed in each case. What are the possible values for $x$ and $u$ in each case?

List some things which would suggest that a certain substitution u(x) could be a useful choice. What do you look for when choosing u(x)?

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