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प्रश्न
\[\int\frac{\sin 2x}{\sin \left( x - \frac{\pi}{6} \right) \sin \left( x + \frac{\pi}{6} \right)} dx\]
योग
उत्तर
\[\text{Let I} = \int\frac{\sin 2x}{\sin\left( x - \frac{\pi}{6} \right) \sin\left( x + \frac{\pi}{6} \right)}dx\]
\[ = \int\frac{\sin 2x}{\sin^2 x - \sin^2 \ sfrac{\pi}{6}} dx \left[ \because \sin \left( A + B \right) \sin\left( A - B \right) = \sin^2 A - \sin^2 B \right]\]
\[ = \int\frac{\sin 2x}{\sin^2 x - \frac{1}{4}}dx\]
\[\text{Putting }\sin^2 x - \frac{1}{4} = t\]
\[ \Rightarrow \text{2}\text{sin x} \text{cos x dx }= dt\]
\[ \Rightarrow \text{sin 2x dx }= dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln }\left| t \right| + C\]
\[ = \text{ln} \left| \sin^2 x - \frac{1}{4} \right| + C \left[ \because t = \sin^2 x - \frac{1}{4} \right]\]
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