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To evaluate the integral 

\[
\int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx
\]

we can simplify the expression by using a substitution. Let’s go through the steps:

1. **Substitution:** Set \( u = \sin x - \cos x \).
   
2. **Differentiate \( u \):**
   \[
   \frac{du}{dx} = \cos x + \sin x \Rightarrow du = (\cos x + \sin x) \, dx
   \]

   This substitution will help because we can rewrite the integral in terms of \( u \) and \( du \).

3. **Rewrite the Integral:**
   Since \( \cos x + \sin x = du/dx \), the integral simplifies to:
   \[
   \int \frac{du}{u} = \ln |u| + C
   \]

4. **Back-Substitute \( u \):**
   Substitute back \( u = \sin x - \cos x \):
   \[
   \ln |\sin x - \cos x| + C
   \]

Thus, the solution to the integral is:

\[
\int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx = \ln |\sin x - \cos x| + C
\]

Would you like further details on any step or have questions?

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Here are 5 related questions that build on this topic:

1. How would you solve an integral with different trigonometric functions in the numerator and denominator?
2. What if the integral had \( \sin x - \cos x \) in the numerator and \( \sin x + \cos x \) in the denominator?
3. How can trigonometric identities simplify integrals with both sine and cosine terms?
4. Why is substitution useful in integrals involving trigonometric expressions?
5. How would this solution change if we were integrating over a definite range?

**Tip:** Recognizing when substitution simplifies an integral, especially with trigonometric forms, can save significant time and effort.

Evaluate the integral \( \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \).

Learn how to evaluate the integral \( \int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx \) using trigonometric substitution. This solution covers the substitution rule and includes step-by-step guidance for simplifying the integral. Suitable for advanced calculus students in Grades 11-12.

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