Question

`  ∫ \sqrt{"cosec x"- 1}  dx `

Answer 1

`  ∫ \sqrt{"cosec x"- 1}  dx `
\[ = \int\sqrt{\frac{1}{\sin x} - 1}dx\]
\[ = \int\frac{\sqrt{1 - \sin x}}{\sqrt{\sin x}}dx\]
\[ = \int\frac{\sqrt{\left( 1 - \sin x \right) \left( 1 + \sin x \right)}}{\sqrt{\sin x \left( 1 + \sin x \right)}}dx\]
\[ = \int\frac{\sqrt{1 - \sin^2 x}}{\sqrt{\sin^2 x + \ sinx}}dx\]
` ∫ {cos  x  dx}/{\sqrt{sin^2 x + sin x}}`
\[\text{Let sin x} = t\]
` ⇒ cos  x   dx = dt  `

Now, `∫  { cos  x  dx }/\sqrt {sin^2  x + sin x} `
\[ = \int\frac{dt}{\sqrt{t^2 + t}}\]
\[ \int\frac{dt}{\sqrt{t^2 + t}}\]
\[ = \int\frac{dt}{\sqrt{t^2 + t + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}\]
\[ = \int\frac{dt}{\sqrt{\left( t + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}\]
\[ = \text{ log }\left| \left( t + \frac{1}{2} \right) + \sqrt{\left( t + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2} \right| + C\]
\[ = \text{ log }\left| t + \frac{1}{2} + \sqrt{t^2 + t} \right| + C\]
\[ = \text{ log }\left| \sin x + \frac{1}{2} + \sqrt{\sin^2 x + \sin x} \right| + C\]