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प्रश्न
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उत्तर
\[\text{ Let I} = \int {cosec}^3 x \text{ dx }\]
\[ = \int {cosec}^2 x \cdot \text{ cosec x dx }\]
\[ = \int {cosec}^2 x \cdot \sqrt{1 + \cot^2 x} \text{ dx }\]
\[\text{ Let} \cot x = t\]
\[ \Rightarrow - {cosec}^2 x \text{ dx } = dt\]
\[ \therefore I = - \int\sqrt{1 + t^2}dt\]
\[ = - \frac{t}{2}\sqrt{1 + t^2} - \frac{1^2}{2} \text{ log} \left| t + \sqrt{1 + t^2} \right| + C . . . (1)\]
\[\text{Substituting the value of t in eq} \text{ (1) }\]
\[ = - \frac{\cot x}{2} \cdot \text{ cosec x }- \frac{1}{2} \text{ log }\left| \text{ cot x + cosec x }\right| + C\]
\[ = - \frac{1}{2}\text{ cosec x cot x} - \frac{1}{2} \text{ log } \left| \frac{\cos x}{\sin x} + \frac{1}{\sin x} \right| + C\]
\[ = - \frac{1}{2} \text{ cosec x cot x }- \frac{1}{2} \text{ log } \left| \frac{2 \cos^2 \frac{x}{2}}{2 \sin \frac{x}{2} \cos \frac{x}{2}} \right| + C\]
\[ = - \frac{1}{2} \text{ cosec x cot x }- \frac{1}{2} \text{ log }\left| \cot \frac{x}{2} \right| + C\]
\[ = - \frac{1}{2} \text{ cosec x cot x} + \frac{1}{2} \text{ log }\left| \tan \frac{x}{2} \right| + C \left( \because \text{ log }\left| \cot \frac{x}{2} \right| = \text{ log }\left| \frac{1}{\tan \frac{x}{2}} \right| \Rightarrow - \text{ log }\left| \tan \frac{x}{2} \right| \right)\]
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