Chapter 5: Q. 12 (page 464)

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

Short Answer

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Ans:

$\sqrt{x^{2} + a^{2}} . . . . . . . . [x = a \tan u] = \sqrt{a^{2} \tan^{2} u + a^{2}} = a s e c u$

Step by step solution

Step 1. Given information:

$x = a \tan u$

A triangle with side lengths a and x and hypotenuse of length $\sqrt{x^{2} + a^{2}}$

Step 2. Solving the trigonometric substitution:

If $x = a \tan u$ , there is no domain restriction on $x$ because $x = a \tan u$ can be any real number. This means that trigonometric substitution $x = a \tan u$ can be used even for integrands that do not have a square roots.

The domain of $x^{2} + a^{2}$ can be any value.

So, the integral involving $x^{2} + a^{2}$ are well suited for trigonometric substitution with $x = a \tan u$ .

Using the Pythagorean identity $a^{2} \tan^{2} u + a^{2} = a^{2} \sec^{2} u$ .

localid="1648777614793" $\sqrt{x^{2} + a^{2}} = \sqrt{a^{2} \tan^{2} u + a^{2}} = \sqrt{a^{2} s e c^{2} u} = |a s e c u| [A . \sqrt{A^{2}} = |A|] = a s e c u$

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

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Solving the trigonometric substitution:

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Explain why using trigonometric substitution with x=atanuoften involves a triangle with side lengths a and x and hypotenuse of lengthx2+a2

Short Answer

Step by step solution

Step 1. Given information:

Step 2. Solving the trigonometric substitution:

One App. One Place for Learning.

Most popular questions from this chapter

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