∫cosect−1dt=∫1−sintsintdt=∫1−sint1+sintsint(1+sint)dt Use =∫1y2−a2=log⁡y+y2−a2+c=∫cos⁡tsin2⁡t+sin⁡tdt Put sin⁡t=y=∫dyy2+y=∫dyy+122−122+c=log⁡y+12+y2+y+c=log⁡sin⁡t+12+sin2⁡t+sin⁡t+c

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$Find \int \sqrt{cosect - 1} d t$

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$\log |\sin t + \sqrt{\sin^{2} t + \sin t}| + c$

$\log |\sin t - \frac{1}{2} + \sqrt{\sin^{2} t + \sin t}| + c$

$\log |\sin t + \frac{1}{2} + \sqrt{\sin^{2} t + \sin t}| + c$

$\log |\sin t + \frac{1}{2} + \sqrt{\sin^{2} t - \sin t}| + c$

answer is A.

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Detailed Solution

$\int \sqrt{\cos e c t - 1} d t = \int \sqrt{\frac{1 - \sin t}{\sin t}} d t$

$= \int \frac{\sqrt{(1 - \sin t) (1 + \sin t)}}{\sqrt{\sin t (1 + \sin t)}} d t$

$Use = \int \frac{1}{\sqrt{y^{2} - a^{2}}} = \log |y + \sqrt{y^{2} - a^{2}}| + c$

$= \int \frac{\cos t}{\sqrt{\sin^{2} t + \sin t}} d t$

$Put \sin t = y$

$\begin{array}{l} = \int \frac{d y}{\sqrt{y^{2} + y}} = \int \frac{d y}{\sqrt{{(y + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}}} + c \\ = \log |(y + \frac{1}{2}) + \sqrt{y^{2} + y}| + c \\ = \log |\sin t + \frac{1}{2} + \sqrt{\sin^{2} t + \sin t}| + c \end{array}$