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प्रश्न
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
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उत्तर
\[\int\frac{\sin \left( 2x \right)}{\left( a + b \cos 2x \right)^2}dx\]
\[\text{Let a + b }\cos2x = t\]
\[ \Rightarrow - \text{b }\sin \left( 2x \right) dx \times 2 = dt\]
\[ \Rightarrow \sin \left( 2x \right) dx = \frac{- dt}{2b}\]
\[Now, \int\frac{\sin \left( 2x \right)}{\left( a + b \cos 2x \right)^2}dx\]
\[ = - \frac{1}{2b}\int\frac{dt}{t^2}\]
\[ = \frac{- 1}{2b}\int t^{- 2} dt\]
\[ = \frac{- 1}{2b}\left[ \frac{t^{- 2 + 1}}{- 2 + 1} \right] + C\]
\[ = \frac{1}{2b} \times \frac{1}{t} + C\]
\[ = \frac{1}{2b \left( a + b \cos 2x \right)} + C\]
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