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Chapter 5: Q. 64 (page 417)

Solve each of the integrals in Exercises 21鈥�70. Some integrals require substitution, and some do not. (Exercise 69 involves a hyperbolic function.)
$鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} d x$

Short Answer

Expert verified

The solution of the given integral is $鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} d x = \frac{1}{2} \sin \sqrt{x} + C$ .

Step by step solution

01

Step 1. Given Information

Solving the given integrals.

$鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} d x$

02

Step 2. Using the substitution method.

Let

$u = \sqrt{x} u = x^{1 / 2} \frac{d u}{d x} = \frac{1}{2} x^{1 / 2 - 1} \frac{d u}{d x} = \frac{1}{2} x^{- 1 / 2} \frac{d u}{d x} = \frac{1}{2} 路 \frac{1}{\sqrt{x}} \frac{1}{2} d u = \frac{1}{\sqrt{x}} d x$

03

Step 3. This substitution changes the integral into

$鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} d x = \frac{1}{2} 鈭� \frac{1}{\sec u} du 鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} dx = \frac{1}{2} 鈭� \cos u du 鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} dx = \frac{1}{2} sinu + C 鈭� \frac{1}{\sqrt{x} \sec \sqrt{x}} dx = \frac{1}{2} \sin \sqrt{x} + C$

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Most popular questions from this chapter

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$x 鈭� 3 x^{2}$

Explain why, if $x = a \tan u$ , then $\sqrt{x^{2} + a^{2}} = a s e c u$ . Your explanation should include a discussion of domains and absolute values.

Explain why, if $x = a s e c u$ , then $\sqrt{a^{2} s e c^{2} u 鈭� a^{2}}$ is $鈭� a \tan u$ if $x < 鈭� a$ and is $a \tan u$ if $x > a$ . Your explanation should include a discussion of domains and absolute values.

Solve $鈭� \frac{x}{4 + x^{2}} d x$ the following two ways:

(a) with the substitution $u = 4 + x^{2};$

(b) with the trigonometric substitution x = 2 tan u.

Solve each of the integrals in Exercises 39鈥�74. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.

$鈭� \frac{\sqrt{4 - x^{2}}}{x^{2}} d x$

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