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Essential Calculus: Early Transcendentals
Found in: Page 326
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 \(\int {{{\sin }^3}\theta {{\cos }^4}} \theta d\theta \)

Transform the integrand in the form \(\left( {{{\cos }^4}\theta - {{\cos }^6}\theta } \right)\sin \theta \)

First of all, we transform the given integrand in the form ‘\(\left( {{{\cos }^4}\theta - {{\cos }^6}\theta } \right)\sin \theta \)’, now we consider \(u = \cos \theta \) followed by its differentiation with respect to \(\theta \) then we substitute the value of ‘\(d\theta \)’ in the integral, on the integration of which we get a general solution. After putting back the value of u in the general solution we obtain the final answer.

See the step by step solution

Step by Step Solution

Step-1:Given data

Given integral is \(\int {{{\sin }^3}\theta {{\cos }^4}\theta d\theta } \)

\(\begin{aligned}{l} &= \int {{{\sin }^2}\theta {{\cos }^4}\theta } d\theta \\ &= \int {\left( {1 - {{\cos }^2}\theta } \right)} {\cos ^4}\theta \sin \theta d\theta \\ &= \int {\left( {{{\cos }^4}\theta - {{\cos }^6}\theta } \right)} \sin \theta d\theta \to (1)\end{aligned}\)

Now let \(u = \cos \theta \) , differentiating ‘\(u\)’ with respect to ‘\(\theta \)’ we get,

\(\frac{{du}}{{d\theta }} = - \sin \theta ,d\theta = - \frac{{du}}{{\sin \theta }}\)

Now, putting the value of \(d\theta \) in (1), we get the integral as

\(\begin{aligned}{l} &= \int {\left( {{u^4} - {u^6}} \right)} \sin \theta \left( { - \frac{1}{{\sin \theta }}} \right)du\\ &= \int { - {u^4}du + \int {{u^6}du} } \\ &= - \int {{u^4}du + \int {{u^6}du} } \end{aligned}\)


On integrating we get,

\(\begin{aligned}{l} - \int {{u^4}du + \int {{u^6}du} } \\\end{aligned}\)

\( = - \frac{1}{5}{u^5} + \frac{1}{7}{u^7} + c\) , where ‘c’ is an arbitrary constant

Now on substituting the value of u, we get the solution as

\( - \frac{1}{5}{\cos ^5}\theta + \frac{1}{7}{\cos ^7}\theta + c\)

Hence, The general solution of the integral \(\int {{{\sin }^3}\theta {{\cos }^4}\theta d\theta } \) is \( - \frac{1}{5}{\cos ^5}\theta + \frac{1}{7}{\cos ^7}\theta + c\) , where ‘c’ is an arbitrary constant

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