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Examples

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Question

$\int \frac{sin x}{sin x - cos x} d x$ is equal to
(where $C$ is integration constant)

\frac{1}{2} ln | sin x - cos x | + \frac{x}{2} + C

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\frac{1}{2} ln | sin x + cos x | + x + C

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\frac{1}{2} ln | sin x + cos x | + \frac{x^{2}}{2} + C

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\frac{1}{2} ln | sin x - cos x | - \frac{x}{2} + C

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Solution

The correct option is A $\frac{1}{2} ln | sin x - cos x | + \frac{x}{2} + C$
$I = \int \frac{sin x}{sin x - cos x} d x = \frac{1}{2} \int \frac{2 sin x}{sin x - cos x} d x = \frac{1}{2} \int \frac{sin x + cos x + sin x - cos x}{sin x - cos x} = \frac{1}{2} \int \frac{cos x + sin x}{sin x - cos x} d x + \frac{1}{2} \int d x = \frac{1}{2} ln | sin x - cos x | + \frac{x}{2} + C$

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