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प्रश्न
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उत्तर
\[\text{ Let I } = \int\frac{\cos x}{\frac{1}{4} - \cos^2 x}dx\]
\[ = \int\frac{\cos x}{\frac{1}{4} - \left( 1 - \sin^2 x \right)}dx\]
\[ = \int\frac{\cos x}{\sin^2 x - \frac{3}{4}}dx\]
\[ = \int\frac{\cos x}{\sin^2 x - \left( \frac{\sqrt{3}}{2} \right)^2}dx\]
\[\text{ Putting sin x = t}\]
\[ \Rightarrow \text{ cos x dx = dt }\]
\[ \therefore I = \int\frac{1}{t^2 - \left( \frac{\sqrt{3}}{2} \right)^2}dt\]
\[ = \frac{1}{2 \times \frac{\sqrt{3}}{2}} \text{ ln }\left| \frac{t - \frac{\sqrt{3}}{2}}{t + \frac{\sqrt{3}}{2}} \right| + C\]
\[ = \frac{1}{\sqrt{3}} \text{ ln } \left| \frac{2t - \sqrt{3}}{2t + \sqrt{3}} \right| + C\]
\[ = \frac{1}{\sqrt{3}} \text{ ln } \left| \frac{2 \sin x - \sqrt{3}}{2 \sin x + \sqrt{3}} \right| + C................ \left[ \because t = \sin x \right]\]
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