`int cot^4(theta) d theta` Use integration tables to find the indefinite integral.

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Indefinite integral follows the formula: `int f(x) dx = F(x)+C`


`f(x)` as the integrand function

`F(x)` as the antiderivative of `f(x) `

`C` as constant of integration.

 The given integral problem: `int cot^4(theta) d theta` resembles one of the formulas from the integration table. It follows the integration formula for cotangent function as :

`int cot^n(x) dx = - (cot^((n-1))(x))/(n-1) - int cot^((n-2)) (x) dx` .

Applying the formula, we get:

`int cot^4(theta) d theta =- (cot^((4-1))(theta))/(4-1) - int cot^((4-2)) (theta) d theta`

                  `=- (cot^3(theta))/3 - int cot^2(theta) d theta`

 To further evaluate the integral part:  `int cot^2(theta) d theta`  we may apply  trigonometric identity: `cot^2(theta) =csc^2(theta) -1` .

`int cot^2(theta) d theta =int [csc^2(theta) -1] d theta`

Apply basic integration property:` int (u-v) dx = int (u) dx - int (v) dx.`

`int [csc^2(theta) -1] d theta =int csc^2(theta) d theta - int 1 d theta`

                                     `= -cot(theta) - theta +C`

Note: From basic integration property: `int dx = x`  then` int 1 d theta = int d theta = theta` .

From the integration table for trigonometric function, we have` int csc^2(x) dx = - cot(x)`  then `int csc^2(theta) d theta=-cot(theta` ).

applying `int [cot^2(theta)] d theta=-cot(theta) - theta +C` , we get the complete indefinite integral as:

`int cot^4(theta) d theta =- (cot^3(theta))/3 - int cot^2(theta) d theta`

                           `=- (cot^3(theta))/3 -(-cot(theta) - theta) +C`

                          `=- (cot^3(theta))/3 + cot(theta) + theta +C`

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