# 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}.

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Given:

• The integral is, {eq}\int {\dfrac{{{{\sin }^3}\theta }}{{\cos \theta - 1}}d\theta } {/eq}.

As it is known that, from the identity,

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