Question:
Evaluate the integral:
\[
\int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx =\ ?
\]
Evaluate the integral:
\[
\int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx =\ ?
\]
Show Hint
When the numerator is the derivative of the denominator, use substitution to convert the integral into a logarithmic form.
Updated On: Jun 4, 2025
- \(-x + \log|\cos x - \sin x| + c\)
- \(x - \log|\cos x - \sin x| + c\)
- \(-\log|\cos x - \sin x| + c\)
- \(\log|\cos x - \sin x| + c\)
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The Correct Option is D
Solution and Explanation
We are given:
\[
\int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx
\]
Let us use substitution. Let:
\[
u = \sin x - \cos x \Rightarrow \frac{du}{dx} = \cos x + \sin x
\]
So,
\[
\frac{\sin x + \cos x}{\sin x - \cos x} \, dx = \frac{du}{u}
\]
Now integrate:
\[
\int \frac{du}{u} = \log|u| + c = \log|\sin x - \cos x| + c
\]
Hence,
\[
\int \frac{\sin x + \cos x}{\sin x - \cos x} \, dx = \boxed{\log|\cos x - \sin x| + c}
\]
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