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Found in: Page 340

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# Verify the formula 31by differential and by substituting $$u = a\sin \theta$$

$$\int {{u^2}\sqrt {{a^2} + {u^2}} } du = \frac{u}{8}\left( {2{u^2} - {a^2}} \right)\sqrt {{a^2} + {u^2}} + \frac{{{a^4}}}{8}{\sin ^{ - 1}}\frac{u}{a} + C$$

See the step by step solution

## Step(1):- Differentiation

$$\frac{u}{8}\left( {2{u^2} - {a^2}} \right)\sqrt {{a^2} + {u^2}} + \frac{{{a^4}}}{8}{\sin ^{ - 1}}\frac{u}{a} + C$$

$$\frac{1}{8}(\left( {2{u^3} - {a^2}u} \right)\sqrt {{a^2} - {u^2}} + {a^4}{\sin ^{ - 1}}\frac{u}{a} + C)$$

## Step(2):- Differentiation by parts:

\begin{aligned}{l}\frac{d}{{du}}((2{u^3} - {a^2}u)\left( {\sqrt {{a^2} + {u^2}} } \right)) = (2{u^3} - {a^2}u)\frac{d}{{du}}\left( {\sqrt {{a^2} + {u^2}} } \right) + \frac{d}{{du}}\left( {(2{u^3} - {a^2}u)} \right)\\ = (2{u^3} - {a^2}u).\frac{1}{{2\sqrt {{a^2} + {u^2}} }}.\left( { - 2u} \right) + 6{u^2} - {a^2}\sqrt {{a^2} + {u^2}} \\ = u\left( {2{u^2} - {a^2}} \right)\frac{u}{{\sqrt {{a^2} + {u^2}} }} + \left( {6{u^2} - {a^2}} \right)\left( {\sqrt {{a^2} + {u^2}} } \right)\\ = \frac{{{u^2}\left( {{a^2} - 2{u^2}} \right)}}{{\sqrt {{a^2} + {u^2}} }} + \left( {6{u^2} - {a^2}} \right)\left( {\sqrt {{a^2} + {u^2}} } \right)\\ = \frac{{{u^2}\left( {{a^2} - 2{u^2}} \right) + \left( {6{u^2} - {a^2}} \right)\left( {\sqrt {{a^2} + {u^2}} } \right)}}{{\sqrt {{a^2} + {u^2}} }}\end{aligned}

## Step(3):- Simplification

We get $$= \frac{{{u^2}\left( {{a^2} - 2{u^2}} \right) + \left( {6{u^2} - {a^2}} \right)\left( {\sqrt {{a^2} + {u^2}} } \right)}}{{\sqrt {{a^2} + {u^2}} }}$$

$$= {u^2}\left( {\sqrt {{a^2} + {u^2}} } \right)$$

Hence proved

## Step(4):- Substitute $$u = a\sin \theta$$

I=$$\int {{u^2}\sqrt {{a^2} + {u^2}} du}$$

=\begin{aligned}{l}u = a\sin \theta \\\frac{{du}}{{d\theta }} = a\cos \theta \\du = a\cos \theta d\theta \\\int {{a^2}{{\sin }^2}\theta \sqrt {{a^2} - {a^2}{{\sin }^2}\theta } a\cos \theta d\theta } \\\int {{a^3}{{\sin }^2}\theta \cos \theta \left( {\sqrt {1 - {{\sin }^2}\theta } } \right)d\theta } \\\int {{a^4}{{\sin }^2}\theta {{\cos }^2}\theta \cos \theta d\theta } \\\int {{a^4}{{\sin }^2}\theta {{\cos }^2}\theta d\theta } \\I = {a^4}\int {{a^4}{{\sin }^2}\theta {{\cos }^2}\theta d\theta } \\ = \frac{{ - {a^4}\left( {\sin \left( {4\theta } \right) - 4\theta } \right)}}{{32}} + C\end{aligned}

We know,

\begin{aligned}{l}u = a\sin \theta \\\sin \theta = \frac{u}{a}\\\theta = {\sin ^{ - 1}}\left( {\frac{u}{a}} \right)\\I = - {a^4}\left( {\sin \left( {4{{\sin }^{ - 1}}\left( {\frac{u}{a}} \right)} \right)} \right) - 4{\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C\end{aligned}

ON simplification, we get

=$$\frac{4}{8}\left( {2{u^2} - {a^2}} \right)\sqrt {{a^2} + {u^2}} + \frac{{{a^4}}}{8}{\sin ^{ - 1}}\frac{u}{a} + C$$

Hence,we get