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