Evaluate the integral:
{eq}\displaystyle \int \frac{sin^3\ \theta}{cos\ \theta - 1}\ d \theta {/eq}
Question:
Evaluate the integral:
{eq}\displaystyle \int \frac{sin^3\ \theta}{cos\ \theta - 1}\ d \theta {/eq}
Indefinite Integral:
The given integral is evaluated by substitution method, the integral is transformed into one form to another form by changing the variable. Consider the function {eq}\int {f\left( x \right)} dx {/eq}, then by assuming {eq}x = g\left( t \right), and \dfrac{{dx}}{{dt}} = g'\left( t \right) {/eq}, or {eq}dx = g'\left( t \right)dt {/eq}. The integral becomes, {eq}\int {f\left( x \right)} dx = \int {f\left[ {g\left( t \right)} \right]} g'\left( t \right)dt {/eq}.
Answer and Explanation: 1
Become a Study.com member to unlock this answer! Create your account
View this answer
Given:
- The integral is, {eq}\int {\dfrac{{{{\sin }^3}\theta }}{{\cos \theta - 1}}d\theta } {/eq}.
As it is known that, from the identity,
...
See full answer below.
Learn more about this topic:
from
Chapter 13 / Lesson 13Learn what integration problems are. Discover how to find integration sums and how to solve integral calculus problems using calculus example problems.