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