Calculus of a Single Variable, Chapter 8, 8.6, Section 8.6, Problem 38

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