Question
Evaluate the following : $\int \frac{1}{1-\sin x} \cdot d x$
✓
Answer
$
\begin{aligned}
\text { I } & =\int\left(\frac{1}{1-\sin x}\right)\left(\frac{1+\sin x}{1+\sin x}\right) \cdot d x \\
& =\int \frac{1+\sin x}{1-\sin ^2 x} \cdot d x \\
& =\int \frac{1+\sin x}{\cos ^2 x} \cdot d x \\
& =\int\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos ^2 x}\right) \cdot d x \\
& =\int\left(\sec ^2 x+\sec x \cdot \tan x\right) \cdot d x \\
& =\tan x+\sec x+c
\end{aligned}
$
\begin{aligned}
\text { I } & =\int\left(\frac{1}{1-\sin x}\right)\left(\frac{1+\sin x}{1+\sin x}\right) \cdot d x \\
& =\int \frac{1+\sin x}{1-\sin ^2 x} \cdot d x \\
& =\int \frac{1+\sin x}{\cos ^2 x} \cdot d x \\
& =\int\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos ^2 x}\right) \cdot d x \\
& =\int\left(\sec ^2 x+\sec x \cdot \tan x\right) \cdot d x \\
& =\tan x+\sec x+c
\end{aligned}
$
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