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Chapter 5: Q 44. (page 452)

Find the integral.
$鈭� \tan^{2} 4 x s e c 4 x d x$

Short Answer

Expert verified

Answer is $\frac{- 1}{8} \ln (|\tan (2 x + \frac{蟺}{4})|) + \frac{1}{8} \tan 4 x s e c 4 x + C$

Step by step solution

01

Step 1. Given information

Integral is $鈭� \tan^{2} 4 x s e c 4 x d x$

02

Step 2. Explanation

Let $4 x = u$

4 d x = d u

$鈭� \tan^{2} 4 x s e c 4 x d x = \frac{1}{4} 鈭� \tan^{2} u s e c u d u = \frac{1}{4} 鈭� (s e c^{2} u - 1) s e c u d u = \frac{1}{4} 鈭� s e c^{3} u - s e c u d u = \frac{- 1}{8} \ln (|\tan (\frac{u}{2} + \frac{蟺}{4})|) + \frac{1}{8} \tan u s e c u + C = \frac{- 1}{8} \ln (|\tan (2 x + \frac{蟺}{4})|) + \frac{1}{8} \tan 4 x s e c 4 x + C$

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Most popular questions from this chapter

Show that if $x = \tan 鈦� u$ , then $d x = \sec^{2} 鈦� u d u$ , in the following two ways: (a) by using implicit differentiation, thinking of $u$ as a function of $x$ , and (b) by thinking of $x$ as a function of $u$ .

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.

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 $x = a s e c u$ ?

For each integral in Exercises 5鈥�8, write down three integrals that will have that form after a substitution of variables.

$鈭� \sin u d u$

Solve given definite integral.

${鈭�}_{4}^{5} 鈥� \frac{1}{x \sqrt{x^{2} + 9}} d x$

See all solutions

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