# `int (theta - csc(theta)cot(theta)) d theta` Find the general indefinite integral.

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### 1 Answer

You need to evaluate the indefinite integral, such that:

`int f(theta)d theta = F(theta) + c`

`int (theta - csc theta* cot theta)d theta = int theta d theta - int (csc theta* cot theta)d theta`

Evaluating integral int theta d theta, using the formula `int theta^n d theta = (theta^(n+1))/(n+1) + c` , yields:

`int theta d theta = (theta^2)/2 + c`

`int (csc theta* cot theta)d theta = int (1/(sin theta)* (cos theta)/(sin theta)) d theta`

You need to use substitution to solve the indefinite integral `int (csc theta* cot theta)d theta` , such that:

`sin theta = t => cos theta d theta = dt`

Replacing the variable, yields:

`int (dt)/(t^2) = int t^(-2) dt = -1/t + c`

Replacing back `sin theta` for t yields:

`int (csc theta* cot theta)d theta = -1/(sin theta) + c`

Gathering the results, yields:

`int (theta - csc theta* cot theta)d theta = (theta^2)/2 + 1/(sin theta) + c`

**Hence, evaluating the indefinite integral yields `int (theta - csc theta* cot theta)d theta = (theta^2)/2 + 1/(sin theta) + c.` **