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Q14E

Expert-verifiedFound in: Page 326

Book edition
2nd

Author(s)
James Stewart

Pages
830 pages

ISBN
9781133112280

**Evaluate the Integral: \(\int {{{\cot }^5}\theta {{\sin }^4}} \theta d\theta \) **

To evaluate this integral, we will use trigonometric identities \(\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}\) and \({\sin ^2}\theta + {\cos ^2}\theta = 1\)

Firstly, we will use \(\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}\) to change the integral in sine and cosine functions. Then we will use \({\sin ^2}\theta + {\cos ^2}\theta = 1\) followed by substitution method

\( = \int {{{\cot }^5}} \theta {\sin ^4}\theta d\theta = \ln |\sin \theta | - {\sin ^2}\theta + \frac{1}{4}{\sin ^4}\theta + c\)

Let \(I = \int {{{\cot }^5}\theta {{\sin }^4}} \theta d\theta \)

As \(\cot \theta = \frac{{\cos \theta }}{{\sin \theta }}\)

\(\begin{aligned}{l}I &= \int {\frac{{{{\cos }^5}\theta }}{{{{\sin }^5}\theta }}} {\sin ^4}\theta d\theta \\ &= \int {\frac{{{{\cos }^5}\theta }}{{\sin \theta }}} d\theta \end{aligned}\)

Split \({\cos ^5}\theta \) as \(\cos \theta {\cos ^4}\theta \) and use \({\cos ^2}\theta = 1 - {\sin ^2}\theta \)

\(\begin{aligned}{l}I &= \int {\frac{{{{\cos }^4}\theta \cos \theta }}{{\sin \theta }}} d\theta \\ &= \int {\frac{{{{(1 - {{\sin }^2}\theta )}^2}}}{{\sin \theta }}} \cos \theta d\theta \end{aligned}\)

Let \(\begin{aligned}{l}\sin \theta &= t\\ &= \cos \theta d\theta &= dt\\I &= \int {\frac{{{{(1 - {t^2})}^2}}}{t}} dt\end{aligned}\)

As \({(a - b)^2} = {a^2} + {b^2} - 2ab\)

\(\begin{aligned}{l}I &= \int {\frac{{{t^4} - 2{t^2}}}{t}} dt\\ &= \int {(\frac{1}{t}} - 2t + {t^3})dt\end{aligned}\)

Split the integral and use \(\int {\frac{1}{\theta }} d\theta = \ln |\theta |\)

\(\begin{aligned}{l}I &= \int {\frac{1}{t}} dt - 2\int {tdt + \int {{t^3}} } dt\\ &= \ln |t| - {t^2} + \frac{{{t^4}}}{4} + c\end{aligned}\)

None Re-Substitute to get the answer

\(I = ln|\sin \theta | - {\sin ^2}\theta + \frac{{{{\sin }^4}\theta }}{4} + c\)

Hence, the value of integral

\(\int {{{\cot }^5}} \theta {\sin ^4}\theta d\theta = \ln |\sin \theta | - {\sin ^2}\theta + \frac{{{{\sin }^4}\theta }}{4} + c\)

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