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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

Expert verified

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

01

Step 1. Given information:

$x = a \tan u$

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

02

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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Most popular questions from this chapter

Suppose v(x) is a function of x. Explain why the integral

of dv is equal to v (up to a constant).

Solve the integral: $鈭� x \ln (x) d x$

Solve given definite integral.

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

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

Give an example of an integral for which trigonometric substitution is possible but an easier method is available. Then give an example of an integral that we still don鈥檛 know how to solve given the techniques we know at this point.

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