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प्रश्न
\[\int\frac{\sin x + \cos x}{\sqrt{\sin 2x}} \text{ dx}\]
योग
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उत्तर
\[\text{ Let I }= \int\left( \frac{\sin x + \cos x}{\sqrt{\sin 2 x}} \right)dx\]
\[\text{ Putting sin x - cos x = t }\]
\[ \Rightarrow \left( \cos x + \sin x \right)dx = dt\]
\[\text{ Also} \left( \text{ sin x} - \cos x \right)^2 = t^2 \]
\[ \Rightarrow \sin^2 x + \cos^2 x - 2 \sin x \cos x = t^2 \]
\[ \Rightarrow 1 - t^2 = \text{ sin }\left( 2x \right)\]
\[ \therefore I = \int\frac{dt}{\sqrt{1 - t^2}}\]
\[ = \sin^{- 1} t + C \left( \int\frac{dt}{\sqrt{a^2 - x^2}} = \sin^{- 1} \frac{x}{a} + C \right)\]
` = \text{ sin}^{- 1} \text{ ( sin x - cos x }) + C ( ∵ t = sin x - cos x ) `
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