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Chapter 5: Q.57 (page 465)

Solve each of the integrals. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.
$鈭� \sqrt{x^{2} - 8 x + 25} d x$

Short Answer

Expert verified

The value is,

$\frac{9}{2} (\sin h^{- 1} (\frac{x - 4}{3})) + \frac{9}{4} \sin h (2 \sin h^{- 1} (\frac{x - 4}{3})) + C$ .

Step by step solution

01

Step 1. Given Information.

The integral is,

$鈭� \sqrt{x^{2} - 8 x + 25} d x$ .

02

Step 2. Simplifying the equation.

$鈭� \sqrt{x^{2} - 8 x + 25} d x = 鈭� \sqrt{{(x - 4)}^{2} + 9} d x = 鈭� \sqrt{u^{2} + 9} d u$

where u=x-4 and du=dx.

Now, let us assume , $u = 3 \sin h (v) d u = 3 \cos h (v) d v$

Now, use the identity, $\sin h^{2} (v) + 1 = co s h^{2} (v)$ .

So,

$\sqrt{u^{2} + 9} = \sqrt{9 \sin h^{2} (v) + 9} = 3 \sqrt{\cos h^{2} (v)}$

03

Step 3. Solving the integral.

$鈭� \sqrt{u^{2} + 9} d u = 鈭� 9 \cos h^{2} (v) d v = 9 鈭� \frac{1}{2} d v + 9 鈭� \cos h^{2} (2 v) d v = \frac{9}{2} (v) + \frac{9}{4} \sin h (w) [w = 2 v, d w = 2 d v] = \frac{9}{2} v + \frac{9}{4} \sin h (2 v) = \frac{9}{2} (\sin h^{- 1} (\frac{u}{3})) + \frac{9}{4} \sin h (2 \sin h^{- 1} (\frac{u}{3})) = \frac{9}{2} (\sin h^{- 1} (\frac{x - 4}{3})) + \frac{9}{4} \sin h (2 \sin h^{- 1} (\frac{x - 4}{3})) + C$

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Most popular questions from this chapter

True/False: Determine whether each of the statements that follow is true or false. If a statement is true, explain why. If a statement is false, provide a counterexample.

(a) True or False: The substitution x = 2 sec u is a suitable choice for solving $鈭� \frac{1}{x^{2} 鈭� 4} d x$ .

(b) True or False: The substitution x = 2 sec u is a suitable choice for solving $鈭� \frac{1}{\sqrt{x^{2} 鈭� 4}} d x$ .

(c) True or False: The substitution x = 2 tan u is a suitable choice for solving $鈭� \frac{1}{x^{2} + 4} d x .$

(d) True or False: The substitution x = 2 sin u is a suitable choice for solving $鈭� {(x^{2} + 4)}^{鈭� 5 / 2} d x$

(e) True or False: Trigonometric substitution is a useful strategy for solving any integral that involves an expression of the form $\sqrt{x^{2} 鈭� a^{2}}$ .

(f) True or False: Trigonometric substitution doesn鈥檛 solve an integral; rather, it helps you rewrite integrals as ones that are easier to solve by other methods.

(g) True or False: When using trigonometric substitution with $x = a \sin u$ , we must consider the cases $x > a$ and $x < - a$ separately.

(h) True or False: When using trigonometric substitution with $x = a s e c u$ , we must consider the cases $x > a$ and $x < - a$ separately.

Solve the integral: $鈭� x \ln (x^{2}) d x$

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 doesn鈥檛 the definite integral ${鈭�}_{2}^{3} \sqrt{1 - x^{2}} d x$ make sense? (Hint: Think about domains.)

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 $x = a s e c u$ ?

See all solutions

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