Explanation Given: a. The intgeral is ∫ x d x 1 − x 2 . b. The intgeral is ∫ x 2 1 − x 2 d x . c. The intgeral is ∫ x 3 1 − x 2 d x . d. The intgeral is ∫ x 4 d x 1 − x 2 . Key concept used: Steps of integration by substitution: If the integral is in the below form: ∫ f ( g ( x ) ) g ′ ( x ) d x Let u = g ( x ) and d u = g ′ ( x ) d x The intgeral becomes, ∫ f ( u ) d u . Thus, integrate f ( u ) . After that substitute g ( x ) back. Reduction formula: ∫ sin m x d x = − cos x sin m − 1 x m + m − 1 m ∫ sin m − 2 x d x Integration formula: ∫ x n d x = x n + 1 n + 1 + c Calculation: Part a: Integration by direct substitution: The intgeral is ∫ x d x 1 − x 2 . ... (1) Let u = 1 − x 2 . Now, differentiate u = 1 − x 2 . d u = − 2 x d x − d u 2 = x d x Substitute u for 1 − x 2 and − d u 2 for x d x in equation (1). ∫ x d x 1 − x 2 = ∫ − 1 2 d u u Simplify the above function. − ∫ d u 2 u = − 1 2 ∫ u − 1 2 d u Integrate the function. − 1 2 ∫ u − 1 2 d u = − 1 2 ⋅ u − 1 2 + 1 − 1 2 + 1 + c = − 1 2 ⋅ u 1 2 1 2 + c = − u + c Now, substitute the value back u = 1 − x 2 . − u + c = − 1 − x 2 + c Hence, the value of ∫ x d x 1 − x 2 is − 1 − x 2 + c . Part b: Integration by trigonometric substitution: The intgeral is ∫ x 2 1 − x 2 d x . ... (2) Let x = sin t . Now, differentiate x = sin t . d x = cos t d t Substitute sin t for x and cos t d t for d x in equation (2). ∫ x 2 1 − x 2 d x = ∫ sin 2 t 1 − sin 2 t cos t d t Simplify the above function. ∫ sin 2 t 1 − sin 2 t cos t d t = ∫ sin 2 t cos t cos t d t = ∫ sin 2 t cos 2 t d t = ∫ sin 2 t ( 1 − sin 2 t ) d t = ∫ sin 2 t d t − ∫ sin 4 t d t Integrate the function. Use the reduction formula: ∫ sin m x d x = − cos x sin m − 1 x m + m − 1 m ∫ sin m − 2 x d x ∫ sin 2 t d t − ∫ sin 4 t d t = ∫ sin 2 t d t − [ − cos t sin 4 − 1 t 4 + 4 − 1 4 ∫ sin 4 − 2 t d t ] = ∫ sin 2 t d t + cos t sin 3 t 4 − 3 4 ∫ sin 2 t d t + c = cos t sin 3 t 4 + 1 4 ∫ sin 2 t d t + c Again, use the reduction formula. cos t sin 3 t 4 + 1 4 ∫ sin 2 t d t + c = cos t sin 3 t 4 + 1 4 [ − sin t cos t 2 + 1 2 ∫ sin 0 t d t ] + c = cos t sin 3 t 4 + 1 4 [ − sin t cos t 2 + 1 2 t ] + c = cos t sin 3 t 4 − sin t cos t 8 + 1 8 t + c Now, substitute the value back x = sin t and t = sin − 1 x . cos t = 1 − sin 2 t = 1 − x 2 cos t sin 3 t 4 − sin t cos t 8 + 1 8 t + c = x 3 1 − x 2 4 − x 1 − x 2 8 + 1 8 sin − 1 x + c Hence, the value of ∫ x 2 1 − x 2 d x is x 3 1 − x 2 4 − x 1 − x 2 8 + 1 8 sin − 1 x + c ...

3rd Edition

ISBN: 9781464125263

Author: Jon Rogawski, Colin Adams

Publisher: W.H. Freeman & Co

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Chapter 8.3, Problem 13E

To determine

a. Evaluate using the substitution $u = 1 - x^{2}$ or by using trigonometric substitution.

b. Evaluate using the substitution $u = 1 - x^{2}$ or by using trigonometric substitution.

c. Evaluate using the substitution $u = 1 - x^{2}$ or by using trigonometric substitution.

d. Evaluate using the substitution $u = 1 - x^{2}$ or by using trigonometric substitution.

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Chapter 8.3, Problem 12E

Chapter 8.3, Problem 14E

Chapter 8 Solutions

Calculus - Standalone book

Ch. 8.1 - Prob. 1PQ Ch. 8.1 - Prob. 2PQ Ch. 8.1 - Prob. 3PQ Ch. 8.1 - Prob. 1E Ch. 8.1 - Prob. 2E Ch. 8.1 - Prob. 3E Ch. 8.1 - Prob. 4E Ch. 8.1 - Prob. 5E Ch. 8.1 - Prob. 6E Ch. 8.1 - Prob. 7E

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