• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration

Q14E

Expert-verified
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

Answers without the blur.

Just sign up for free and you're in.

Illustration

Short Answer

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

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

See the step by step solution

Step by Step Solution

Step 1: Using Trigonometric Identities

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

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

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

Step 2: Splitting of fifth power of cosine function

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

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

Step 3: Substitution Method

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

Step 4: Using Square identity formula

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

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

Step 5: Splitting Of Integral

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

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

Step 6: Re-Substitution of the value of t

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

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

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.