Book edition 1st

Author(s) Peter Kohn, Laura Taalman

Pages 1155 pages

ISBN 9781429241861

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Short Answer

Solve the integral $\int x^{3} \sqrt{x^{2} - 1} d x$ three ways:
(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;
(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$
(c) with the trigonometric substitution x = sec u.

Part (a) The solution of the integral is $\frac{1}{15} ({(x^{2} - 1)}^{\frac{3}{2}} (3 x^{2} + 2)) + C .$

Part (b) The solution of the integral is $\frac{1}{15} ({(x^{2} - 1)}^{\frac{3}{2}} (3 x^{2} + 2)) + C .$

Part (c) The solution of the integral is $\frac{1}{15} ({(x^{2} - 1)}^{\frac{3}{2}} (3 x^{2} + 2)) + C .$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Part (a) Step 1. Given Information.

The given integral is $\int x^{3} \sqrt{x^{2} - 1} d x .$

Part (a) Step 2. Solve.

We have to solve the integral with the substitution $u = x^{2} - 1,$ so the derivative is $d u = 2 x d x .$

Let's solve the integral by substituting u,

localid="1648816067865" $\int x^{3} \sqrt{x^{2} - 1} d x = \frac{1}{2} \int (u + 1) \sqrt{u} d u = \frac{1}{2} \int u^{\frac{3}{2}} + \sqrt{u} d u = \frac{1}{2} \int u^{\frac{3}{2}} d u + \frac{1}{2} \int \sqrt{u} d u = \frac{1}{2} [\frac{2}{5} u^{\frac{5}{2}}] + \frac{1}{2} [\frac{2}{3} u^{\frac{3}{2}}] + C = \frac{1}{5} u^{\frac{5}{2}} + \frac{1}{3} u^{\frac{3}{2}} + C Substitute back u, = \frac{1}{5} {(x^{2} - 1)}^{\frac{5}{2}} + \frac{1}{3} {(x^{2} - 1)}^{\frac{3}{2}} + C = \frac{1}{15} ({(x^{2} - 1)}^{\frac{3}{2}} (3 x^{2} + 2)) + C$

Part (b) Step 1. Solve.

We have to solve the integral with integration by parts, choosing $u = x^{2}$ and $d v = x \sqrt{x^{2} - 1} d x .$

Thus, the derivation of u is $d u = (2 x) d x,$ and to find v integrate dv.

So,

role="math" localid="1648817215834" $\int d v = \int x \sqrt{x^{2} - 1} d x v = \frac{1}{3} {(x^{2} - 1)}^{\frac{3}{2}} d x$

Part (b) Step 2. Solve.

Let's do the integration by parts,

$\int x^{2} (x \sqrt{x^{2} - 1}) d x = x^{2} (\frac{1}{3} {(x^{2} - 1)}^{\frac{3}{2}}) - \frac{1}{3} \int {(x^{2} - 1)}^{\frac{3}{2}} (2 x d x) = \frac{x^{2}}{3} {(x^{2} - 1)}^{\frac{3}{2}} - \frac{2}{3} \int {(x^{2} - 1)}^{\frac{3}{2}} (x d x) = \frac{x^{2}}{3} {(x^{2} - 1)}^{\frac{3}{2}} - \frac{2}{3} [\frac{1}{5} {(x^{2} - 1)}^{\frac{5}{2}}] + C = \frac{1}{15} {(x^{2} - 1)}^{\frac{3}{2}} (5 x^{2} - 2 (x^{2} - 1)) + c = \frac{1}{15} {(x^{2} - 1)}^{\frac{3}{2}} (3 x^{2} + 2) + C$

Part (c) Step 1. Solve.

We have to solve the integral with the substitution $x = s e c u,$ so the derivative is $d x = s e c u \tan u d u .$

Let's solve the integral by substituting x,

role="math" localid="1648817837174" $\int x^{3} \sqrt{x^{2} - 1} d x = \int s e c^{3} u \sqrt{s e c^{2} u - 1} (s e c u \tan u) d u = \int s e c^{4} u \sqrt{\tan^{2} u} (\tan u) d u = \int s e c^{4} u \tan^{2} u d u = \int s e c^{2} u s e c^{2} u \tan^{2} u d u = \int (1 + t {an}^{2} u) s e c^{2} u \tan^{2} u d u Let t = t a n u, d t = s e c^{2} u d u = \int (1 + t^{2}) t^{2} d t = \int t^{2} + t^{4} d t = \int t^{2} d t + \int t^{4} d t = \frac{t^{3}}{3} + \frac{t^{5}}{5} + C$

Part (c) Step 2. Solve.

By proceeding with the calculation further,

$Substitute back t, = \frac{{(tanu)}^{3}}{3} + \frac{{(tanu)}^{5}}{5} + C = \frac{{(\sqrt{\sec^{2} u - 1})}^{3}}{3} + \frac{{(\sqrt{\sec^{2} u - 1})}^{5}}{5} + C = \frac{{(\sec^{2} u - 1)}^{\frac{3}{2}}}{3} + \frac{{(\sec^{2} u - 1)}^{\frac{5}{2}}}{5} + C Substitute back u, = \frac{{(x^{2} - 1)}^{\frac{3}{2}}}{3} + \frac{{(x^{2} - 1)}^{\frac{5}{2}}}{5} + C = \frac{1}{15} ({(x^{2} - 1)}^{\frac{3}{2}} (3 x^{2} + 2)) + C$

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Calculus

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Short Answer

Solve the integral $\int x^{3} \sqrt{x^{2} - 1} d x$ three ways:
(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;
(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$
(c) with the trigonometric substitution x = sec u.

Step by Step Solution

Part (a) Step 1. Given Information.

Part (a) Step 2. Solve.

Part (b) Step 1. Solve.

Part (b) Step 2. Solve.

Part (c) Step 1. Solve.

Part (c) Step 2. Solve.

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Solve the integral∫x3x2-1dx three ways:(a) with the substitution u=x2-1, followed by back substitution;(b) with integration by parts, choosing localid="1648814744993" u=x2 and dv=xx2-1dx;(c) with the trigonometric substitution x = sec u.

Step by Step Solution

Part (a) Step 1. Given Information.

Part (a) Step 2. Solve.

Part (b) Step 1. Solve.

Part (b) Step 2. Solve.

Part (c) Step 1. Solve.

Part (c) Step 2. Solve.

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Solve the integral $\int x^{3} \sqrt{x^{2} - 1} d x$ three ways:
(a) with the substitution $u = x^{2} - 1,$ followed by back substitution;
(b) with integration by parts, choosing localid="1648814744993" $u = x^{2} and d v = x \sqrt{x^{2} - 1} d x;$
(c) with the trigonometric substitution x = sec u.