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Question
Write a value of \[\int\frac{1 - \sin x}{\cos^2 x} \text{ dx }\]
Solution
\[\int\left( \frac{1 - \sin x}{\cos^2 x} \right) dx\]
\[ = \int\left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos^2 x} \right)dx\]
\[ \int\left( \frac{1}{\cos^2 x} - \frac{\sin x}{\cos x} \times \frac{1}{\cos x} \right) dx\]
\[ = \int\left( \sec^2 x - \sec x \tan x \right) dx\]
\[ = \tan x - \sec x + C\]
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