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Question
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Solution
\[\int\frac{1}{\sin x \cos^2 x}dx\]
\[ = \int\frac{\sin^2 x + \cos^2 x}{\sin x \cos^2 x}dx\]
\[ = \int\tan x \sec x + cosec\ x\ dx\]
\[ = \sec x + \text{ln} \left| cosec\ x - \cot x \right| + C\]
\[ = \sec x + \text{ln} \left| \tan\frac{x}{2} \right| + C \left[ \because cosec\ x - \ cot\ x = \frac{1 - \ cosx}{\sin x} = \tan\frac{x}{2} \right]\]
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