Chapter 5: Q. 94 (page 466)

Prove part (b) of Theorem 5.18: For $x \in (- \infty, \infty) and u \in (- π / 2, π / 2),$ the substitution $x = a \tan u$ gives $x^{2} + a^{2} = a \sec u$ . Your proof should include a discussion of domains and a consideration of absolute values.

Short Answer

Expert verified

It is proved.

Step by step solution

Step 1. Given information.

It is given $x \in (- \infty, \infty), u \in (- π / 2, π / 2)$

Step 2. Explanation.

For $x = a \tan u$

We know, From Pythagorean Theorem,

$a^{2} \tan^{2} u + a^{2} = a^{2} \sec^{2} u \sqrt{x^{2} + a^{2}} = \sqrt{a^{2} \tan^{2} u + a^{2}} = \sqrt{a^{2} \sec^{2} u} = | a \sec u | = a \sec u$

Hence proved.

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Prove part (b) of Theorem 5.18: For $x \in (- \infty, \infty) and u \in (- π / 2, π / 2),$ the substitution $x = a \tan u$ gives $x^{2} + a^{2} = a \sec u$ . Your proof should include a discussion of domains and a consideration of absolute values.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Explanation.

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Prove part (b) of Theorem 5.18: Forx∈(-∞,∞)andu∈(-π/2,π/2),the substitution x=atanu gives x2+a2=asecu. Your proof should include a discussion of domains and a consideration of absolute values.

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Explanation.

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Prove part (b) of Theorem 5.18: For $x \in (- \infty, \infty) and u \in (- π / 2, π / 2),$ the substitution $x = a \tan u$ gives $x^{2} + a^{2} = a \sec u$ . Your proof should include a discussion of domains and a consideration of absolute values.