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प्रश्न
\[\int\frac{1 - \sin 2x}{x + \cos^2 x} dx\]
बेरीज
उत्तर
\[\text{Let I} = \int\frac{1 - \sin2x}{x + \cos^2 x}dx\]
\[\text{Putting}\ x + \cos^2 x = t\]
\[ \Rightarrow 1 - 2\ cosx . \ sinx = \frac{dt}{dx}\]
\[ \Rightarrow \left( 1 - \sin 2x \right)dx = dt\]
\[ \therefore I = \int\frac{1}{t}dt\]
\[ = \text{ln} \left| t \right| + C\]
\[ = \text{ln }\left| x + \cos^2 x \right| + C \left[ \because t = x + \cos^2 x \right]\]
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