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