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प्रश्न
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
योग
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उत्तर
\[\text{Let I} = \int\frac{- \sin x + 2\cos x}{2\sin x + \cos x}dx\]
\[\text{Putting}\ 2\sin x + \cos x = t\]
\[ \Rightarrow 2\cos x - \sin x = \frac{dt}{dx}\]
\[ \Rightarrow \left( - \sin x + 2\cos x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln}\left| t \right| + C\]
\[ = \text{ln }\left| 2\sin x + \cos x \right| + C \left[ \because t = 2\sin x + \cos x \right]\]
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