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