The given function is \( (\tan x + \cot x)^2 \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \( \int (\tan x + \cot x)^2 \, dx \).
Solution:
First, expand \( (\tan x + \cot x) ^2 \):
\( \begin{aligned} &(\tan x + \cot x)^2 \\ &= \tan^2 x + 2 \tan x \cot x + \cot^2 x \\ &= \tan^2 x + 2 \cdot 1 + \cot^2 x, \quad \because \tan x \cot x = 1 \\ &= \sec^2 x \,- 1 + \csc^2 x \,- 1 + 2, \quad \because \tan^2 x = \sec^2 x – 1, \cot^2 x = \csc^2 x – 1 \\ &= \sec^2 x + \csc^2 x \\ \end{aligned} \)So the integral becomes:
\( \begin{aligned} &\int (\sec^2 x + \csc^2 x) \, dx \\ &= \int \sec^2 x \, dx + \int \csc^2 x \, dx \\ &= \tan x \, – \cot x + C, \quad \text{C is a constant} \\ &= \frac{\sin x}{\cos x} – \frac{\cos x}{\sin x} + C \\ &= \frac{\sin^2 x \, – \cos^2 x}{\sin x \cos x} + C \\ &= \frac{-\cos 2x}{\frac{\sin 2x}{2}} + C\\ &= -2 \cot 2x + C \\ \end{aligned} \)Thus, the integral of the given function is
\( \boxed{ \int (\tan x + \cot x)^2 \, dx = -2 \cot 2x + C} \\ \)Please let me know in the comments if you find any errors in this solution.