Question

\( \int e^x (1 - \cot(x) + \cot^2(x)) \, dx = ? \)

Answer 1

Step 1: Simplify the Integrand using Trigonometric Identities
We are given the integral: \[ \int e^x (1 - \cot(x) + \cot^2(x)) \, dx \] Recall that: \[ \cot^2(x) = \csc^2(x) - 1 \] Substituting this identity into the integral, we get: \[ \int e^x \left( 1 - \cot(x) + \csc^2(x) - 1 \right) \, dx \] Simplifying the expression inside the integrand: \[ \int e^x (\csc^2(x) - \cot(x)) \, dx \]

Step 2: Recognize the Derivative Relationship
We know that the derivative of \( \cot(x) \) with respect to \( x \) is \( -\csc^2(x) \), i.e.: \[ \frac{d}{dx} \cot(x) = -\csc^2(x) \] Let \( f(x) = \cot(x) \), so that \( f'(x) = -\csc^2(x) \). Therefore, \( \csc^2(x) = -f'(x) \).

Step 3: Rewrite the Integral in Terms of \( f(x) \) and \( f'(x) \)
Substituting \( f(x) \) and \( f'(x) \) into the integral, we get: \[ \int e^x (\csc^2(x) - \cot(x)) \, dx = \int e^x \left( -f'(x) - f(x) \right) \, dx \] This can be written as: \[ - \int e^x (f(x) + f'(x)) \, dx \]

Step 4: Apply the Integration Rule for \( e^x [f(x) + f'(x)] \)
We apply the standard rule for integrating expressions of the form \( \int e^x (f(x) + f'(x)) \, dx \), which states: \[ \int e^x (f(x) + f'(x)) \, dx = e^x f(x) + C \] Since our integral has a negative sign: \[ - \int e^x (f(x) + f'(x)) \, dx = -e^x f(x) + C \]

Step 5: Substitute \( f(x) = \cot(x) \) to Obtain the Final Result
Now, substitute \( f(x) = \cot(x) \) back into the expression: \[ -e^x f(x) + C = -e^x \cot(x) + C \]

Therefore, the integral is:
\[ \int e^x (1 - \cot(x) + \cot^2(x)) \, dx = -e^x \cot(x) + C \]