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प्रश्न
\[\int\frac{\cos x}{2 + 3 \sin x} dx\]
योग
उत्तर
\[\text{Let I} = \int\frac{\cos x}{2 + 3\sin x}dx\]
\[\text{Putting }\sin x = t \]
\[ \Rightarrow \cos x = \frac{dt}{dx}\]
\[ \Rightarrow \text{cos x dx} = dt\]
\[ \therefore I = \int\frac{dt}{2 + 3t}\]
\[ = \frac{1}{3}\text{ln }\left| 2 + 3t \right| + C \left[ \because \int\frac{1}{ax + b}dx = \frac{1}{a}\text{ln }\left| ax + b \right| + C \right]\]
\[ = \frac{1}{3} \text{ln} \left| 2 + 3 \sin x \right| + C \left[ \because t = \sin x \right]\]
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