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Q. 38

Expert-verified
Found in: Page 464

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

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

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

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

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

See the step by step solution

Part (a) Step 1. Given Information.

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

Part (a) Step 2. Solve.

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

Let's solve the integral by substituting u,

localid="1648816067865" $\int {x}^{3}\sqrt{{x}^{2}-1}dx\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int \left(u+1\right)\sqrt{u}du\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int {u}^{3}{2}}+\sqrt{u}du\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\int {u}^{3}{2}}du+\frac{1}{2}\int \sqrt{u}du\phantom{\rule{0ex}{0ex}}=\frac{1}{2}\left[\frac{2}{5}{u}^{5}{2}}\right]+\frac{1}{2}\left[\frac{2}{3}{u}^{3}{2}}\right]+C\phantom{\rule{0ex}{0ex}}=\frac{1}{5}{u}^{5}{2}}+\frac{1}{3}{u}^{3}{2}}+C\phantom{\rule{0ex}{0ex}}\mathrm{Substitute}\mathrm{back}u,\phantom{\rule{0ex}{0ex}}=\frac{1}{5}{\left({x}^{2}-1\right)}^{5}{2}}+\frac{1}{3}{\left({x}^{2}-1\right)}^{3}{2}}+C\phantom{\rule{0ex}{0ex}}=\frac{1}{15}\left({\left({x}^{2}-1\right)}^{3}{2}}\left(3{x}^{2}+2\right)\right)+C$

Part (b) Step 1. Solve.

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

Thus, the derivation of u is $du=\left(2x\right)dx,$ and to find v integrate dv.

So,

role="math" localid="1648817215834" $\int dv=\int x\sqrt{{x}^{2}-1}dx\phantom{\rule{0ex}{0ex}}v=\frac{1}{3}{\left({x}^{2}-1\right)}^{3}{2}}dx$

Part (b) Step 2. Solve.

Let's do the integration by parts,

$\int {x}^{2}\left(x\sqrt{{x}^{2}-1}\right)dx\phantom{\rule{0ex}{0ex}}={x}^{2}\left(\frac{1}{3}{\left({x}^{2}-1\right)}^{3}{2}}\right)-\frac{1}{3}\int {\left({x}^{2}-1\right)}^{3}{2}}\left(2xdx\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{3}{\left({x}^{2}-1\right)}^{3}{2}}-\frac{2}{3}\int {\left({x}^{2}-1\right)}^{3}{2}}\left(xdx\right)\phantom{\rule{0ex}{0ex}}=\frac{{x}^{2}}{3}{\left({x}^{2}-1\right)}^{3}{2}}-\frac{2}{3}\left[\frac{1}{5}{\left({x}^{2}-1\right)}^{5}{2}}\right]+C\phantom{\rule{0ex}{0ex}}=\frac{1}{15}{\left({x}^{2}-1\right)}^{3}{2}}\left(5{x}^{2}-2\left({x}^{2}-1\right)\right)+c\phantom{\rule{0ex}{0ex}}=\frac{1}{15}{\left({x}^{2}-1\right)}^{3}{2}}\left(3{x}^{2}+2\right)+C$

Part (c) Step 1. Solve.

We have to solve the integral with the substitution $x=secu,$ so the derivative is $dx=secu\mathrm{tan}udu.$

Let's solve the integral by substituting x,

role="math" localid="1648817837174" $\int {x}^{3}\sqrt{{x}^{2}-1}dx\phantom{\rule{0ex}{0ex}}=\int se{c}^{3}u\sqrt{se{c}^{2}u-1}\left(secu\mathrm{tan}u\right)du\phantom{\rule{0ex}{0ex}}=\int se{c}^{4}u\sqrt{{\mathrm{tan}}^{2}u}\left(\mathrm{tan}u\right)du\phantom{\rule{0ex}{0ex}}=\int se{c}^{4}u{\mathrm{tan}}^{2}udu\phantom{\rule{0ex}{0ex}}=\int se{c}^{2}use{c}^{2}u{\mathrm{tan}}^{2}udu\phantom{\rule{0ex}{0ex}}=\int \left(1+t{\mathrm{an}}^{2}u\right)se{c}^{2}u{\mathrm{tan}}^{2}udu\phantom{\rule{0ex}{0ex}}\mathrm{Let}t=tanu,dt=se{c}^{2}udu\phantom{\rule{0ex}{0ex}}=\int \left(1+{t}^{2}\right){t}^{2}dt\phantom{\rule{0ex}{0ex}}=\int {t}^{2}+{t}^{4}dt\phantom{\rule{0ex}{0ex}}=\int {t}^{2}dt+\int {t}^{4}dt\phantom{\rule{0ex}{0ex}}=\frac{{t}^{3}}{3}+\frac{{t}^{5}}{5}+C$

Part (c) Step 2. Solve.

By proceeding with the calculation further,