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Chapter 5: Q. 91 (page 420)

Prove the integration formula
$鈭� \tan x d x = \ln | \sec x | + C$
(a) by using algebra and integration by substitution to find tan x dx;
(b) by differentiating ln | sec x|.

Short Answer

Expert verified

(a) After using algebra and integration by substitution we prove that $鈭� tanxdx = \ln (\sec) + C$

(b) After differentiating we prove that $\frac{d}{d x} \ln | \sec x | = \tan x$

Step by step solution

01

Step 1. Given Information

Prove the integration formula

$鈭� \tan x d x = \ln | \sec x | + C$

(a) by using algebra and integration by substitution to find $鈭� \tan x d x$ ;

(b) by differentiating $\ln | \sec x |$ .

02

Part (a) Step 1. Using algebra and integration by substitution to find ∫tanxdx.

We can write as

$鈭� \tan x d x = 鈭� \frac{\sin x}{\cos x} d x$

Let

$u = \cos x \frac{d u}{d x} = - \sin x d u = - \sin x d x - d u = \sin x d x$

03

Part (a) Step 2. Now the integral after substitution.

$鈭� \tan x d x = - 鈭� \frac{1}{u} du 鈭� tanxdx = - \ln (u) + C 鈭� tanxdx = - \ln (cosx) + C 鈭� tanxdx = \ln (\sec x) + C$

04

Part (b) Step 1. By differentiating ln|secx|.

$\frac{d}{d x} \ln | \sec x | = \frac{d}{d x} \ln | \sec x | 路 \frac{d}{d x} \sec x \frac{d}{d x} \ln | \sec x | = \frac{1}{\sec x} 路 \sec x \tan x \frac{d}{d x} \ln | \sec x | = \tan x$

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Most popular questions from this chapter

Find three integrals in Exercises 39鈥�74 that can be solved by using a trigonometric substitution of the form $x = a \tan u$ .

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.

$鈭� x^{3} \sqrt{x^{2} + 1} d x$

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.

Explain why using trigonometric substitution with $x = a \tan u$ often involves a triangle with side lengths a and x and hypotenuse of length $\sqrt{x^{2} + a^{2}}$

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" $(c) 鈭� \frac{x^{2}}{4 + x^{2}} d x$ $(d) 鈭� \frac{16 鈭� x^{4}}{4 + x^{2}} d x$

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