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प्रश्न
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उत्तर
\[\text{ Let I} = \int\frac{\sin x}{\text{ cos 2 x}} dx\]
\[ = \int\left( \frac{\sin x}{2 \cos^2 x - 1} \right) dx ...................\left[ \because \cos 2x = 2 \cos^2 x - 1 \right] \]
\[\text{ Putting cos x = t}\]
\[ \Rightarrow - \text{ sin x dx = dt}\]
\[ \Rightarrow \text{ sin x dx = - dt} \]
\[ \therefore I = \int\frac{- dt}{2 t^2 - 1}\]
\[ = \frac{1}{2}\int\frac{- dt}{t^2 - \frac{1}{2}}\]
\[ = \frac{- 1}{2}\int\frac{dt}{t^2 - \left( \frac{1}{\sqrt{2}} \right)^2}\]
` = -1 / 2 × 1/ 2 × 1/1\sqrt2 In |{t -1/\sqrt2}/{t+1/\sqrt2 }| + C ..... [ ∵ ∫ {1}/{x^2 - a^2} = {1}/{2a}\text{ ln } | {x - a}/{x + a} | + C ] `
\[ = - \frac{1}{2\sqrt{2}} \text{ ln} \left| \frac{\sqrt{2}t - 1}{\sqrt{2}t + 1} \right| + C\]
\[ = - \frac{1}{2\sqrt{2}} \text{ ln }\left| \frac{\sqrt{2} \cos x - 1}{\sqrt{2} \cos x + 1} \right| + C ............\left[ \because t = \cos x \right]\]
\[ = \frac{1}{2\sqrt{2}} \text{ ln} \left| \frac{\sqrt{2} \cos x + 1}{\sqrt{2} \cos x - 1} \right| + C\]
