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Essential Calculus: Early Transcendentals
Found in: Page 306
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

Evaluate the integral by making the given substitution.

\(\int {{{\cos }^3}} \theta \sin \theta d\theta ,\;\;\;u = \cos \theta \)

The value of \(\int {{{\cos }^3}} \theta \sin \theta d\theta \)is\( - \frac{{{{(\cos \theta )}^4}}}{4} + C\).

See the step by step solution

Step by Step Solution

Step 1 Definition of the integral

Integral calculus is an area of mathematics that deals with the calculation, properties, and applications of integrals.

Step 2 : Evaluating the integral.                                        

The given data is expressed as,

\(\int {{{\cos }^3}} \theta \sin \theta d\theta \)

After Substituting

\(\cos \theta = u\)


\(\begin{aligned}{c} - \sin \theta d\theta &= du\\\sin \theta d\theta &= - du\end{aligned}\)

\(\begin{aligned}{c}\int {{{\cos }^3}} \theta \sin \theta d\theta &= - \int {{u^3}} du\\ &= - \frac{{{u^4}}}{4} + C\end{aligned}\)

Substituting again

\(u = \cos \theta \),

We get

\(\int {{{\cos }^3}} \theta \sin \theta d\theta = - \frac{{{{(\cos \theta )}^4}}}{4} + C\)

Hence the value of \(\int {{{\cos }^3}} \theta \sin \theta d\theta \)is\( - \frac{{{{(\cos \theta )}^4}}}{4} + C\).

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