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प्रश्न
\[\int\frac{1 - \sin x}{x + \cos x} dx\]
बेरीज
उत्तर
\[\text{Let I} = \int\frac{1 - \sin x}{x + \cos x}dx\]
\[\text{Putting x} + \cos x = t\]
\[ \Rightarrow 1 - \sin x = \frac{dt}{dx}\]
\[ \Rightarrow \left( 1 - \sin x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln t} + C\]
\[ = \text{ln }\left| x + \cos x \right| + C \left[ \because t = x + \cos x \right]\]
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