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Chapter 5: Q. 61 (page 465)

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

Short Answer

Expert verified

The value is,

$\frac{x}{2} \sqrt{- x^{2} + 2} + \sin^{- 1} (\frac{\sqrt{2} x}{2}) + C$

Step by step solution

01

Step 1. Given Information.

The integral is,

$鈭� \sqrt{2 - x^{2}} d x$

02

Step 2. Simplifying the integral.

Let $x = \sqrt{2} \sin (u)$

Now, The derivation is

$x = \sqrt{2} \sin (u) d x = \sqrt{2} \cos (u) d u$

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

Now,

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

03

Step 3. Solving the integral.

The integral is,

$鈭� \sqrt{2 - x^{2}} d x = 鈭� 2 \cos^{2} (u) d u = 2 鈭� (\frac{1}{2} \cos (2 u)) d u + 鈭� d u = u + 鈭� \cos (2 u) d u = u + \frac{1}{2} 鈭� \cos (v) d v [v = 2 u, d v = 2 d u] = u + \frac{1}{2} \sin (v) + C = u + \frac{1}{2} \sin (2 u) + C = \sin^{- 1} (\frac{\sqrt{2} x}{2}) + \frac{1}{2} \sin (2 \sin^{- 1} (\frac{\sqrt{2} x}{2})) + C = \frac{x}{2} \sqrt{- x^{2} + 2} + \sin^{- 1} (\frac{\sqrt{2} x}{2}) + C$

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Most popular questions from this chapter

Solve given definite integral.

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

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