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Chapter 5: Q. 64 (page 465)

Solve each of the integrals. Some integrals require trigonometric substitution, and some do not. Write your answers as algebraic functions whenever possible.
$鈭� \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x$

Short Answer

Expert verified

The value is, $- \frac{\sqrt{- x^{2} + 4}}{4 x} + C$ .

Step by step solution

01

Step 1. Given Information.

The integral is,

$鈭� \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x$ .

02

Step 2. Simplifying the integral.

Let $x = 2 \sin (u)$ .

Now, the derivation of the integral is,

$x = 2 \sin (u) d x = 2 \cos (u) d u$

Using the identity, $- \sin^{2} (u) + 1 = \cos^{2} (u)$ ,

Now,

$\frac{1}{x^{2} \sqrt{4 - x^{2}}} = \frac{1}{4 \sin^{2} (u) \sqrt{4 - 4 \sin^{2} (u)}} = \frac{1}{8 \sin^{2} (u) \sqrt{1 - \sin^{2} (u)}} = \frac{1}{8 \sin^{2} (u) \sqrt{\cos^{2} (u)}} = \frac{1}{8 \sin^{2} (u) \cos (u)}$

03

Step 3. Solving the integral.

The solution of the integral is,

$鈭� \frac{1}{x^{2} \sqrt{4 - x^{2}}} d x = 鈭� \frac{1}{4 \sin^{2} (u)} d u = \frac{1}{4} 鈭� c s c^{2} (u) d u = \frac{1}{4} (- c o t (u)) + C = \frac{1}{4} (- c o t (\sin^{- 1} (\frac{x}{2}))) + C = \frac{\sqrt{- \frac{x^{2}}{4} + 1}}{2 x} + C = \frac{\sqrt{- x^{2} + 4}}{4 x} + C$

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Most popular questions from this chapter

For each function u(x) in Exercises 9鈥�12, write the differential du in terms of the differential dx.

$u (x) = \sin x$

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.

Solve the integral: $鈭� \frac{3 x + 1}{s e c x} d x$

Complete the square for each quadratic in Exercises 28鈥�33. Then describe the trigonometric substitution that would be appropriate if you were solving an integral that involved that quadratic.

$2 x^{2} 鈭� 4 x + 1$

Solve the integral: $鈭� \ln \sqrt{x} d x$

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