Question
Evaluate : $\int \frac{x}{1-\sin x} \cdot d x$
✓
Answer
$
\begin{aligned}
& \mathrm{I}=\int \frac{x}{1-\sin x} \cdot \frac{(1+\sin x)}{(1+\sin x)} \cdot d x \\
& =\int \frac{x(1+\sin x)}{1-\sin ^2 x} \cdot d x=\int \frac{x(1+\sin x)}{\cos ^2 x} \cdot d x=\int x \cdot\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos ^2 x}\right) \cdot d x \\
& =\int x \cdot\left(\sec ^2 x+\sec x \cdot \tan x\right) \cdot d x \\
& =\int x \cdot \sec ^2 x \cdot d x+\int x \cdot \sec x \cdot \tan x \cdot d x \\
& =\left(x \cdot \int \sec ^2 x \cdot d x-\int \frac{d}{d x} x \cdot \int \sec ^2 x \cdot d x \cdot d x\right)+\left(x \cdot \int \sec x \cdot \tan x \cdot d x-\int \frac{d}{d x} x \cdot \int \sec x \cdot \tan x \cdot d x \cdot d x\right) \\
& =x \cdot \tan x-\int(1) \cdot \tan x \cdot d x+x \cdot \sec x-\int(1) \cdot \sec x \cdot d x \\
& =x \cdot \tan x-\log (\sec x)+x \cdot \sec x-\log (\sec x+\tan x)+c \\
& =x \cdot(\sec x+\tan x)-\log (\sec x)-\log (\sec x+\tan x)+c \\
& \therefore \quad \int \frac{x}{1-\sin x} \cdot d x=x \cdot(\sec x+\tan x)-\log [(\sec x)(\sec x+\tan x)]+c \\
\end{aligned}
$
\begin{aligned}
& \mathrm{I}=\int \frac{x}{1-\sin x} \cdot \frac{(1+\sin x)}{(1+\sin x)} \cdot d x \\
& =\int \frac{x(1+\sin x)}{1-\sin ^2 x} \cdot d x=\int \frac{x(1+\sin x)}{\cos ^2 x} \cdot d x=\int x \cdot\left(\frac{1}{\cos ^2 x}+\frac{\sin x}{\cos ^2 x}\right) \cdot d x \\
& =\int x \cdot\left(\sec ^2 x+\sec x \cdot \tan x\right) \cdot d x \\
& =\int x \cdot \sec ^2 x \cdot d x+\int x \cdot \sec x \cdot \tan x \cdot d x \\
& =\left(x \cdot \int \sec ^2 x \cdot d x-\int \frac{d}{d x} x \cdot \int \sec ^2 x \cdot d x \cdot d x\right)+\left(x \cdot \int \sec x \cdot \tan x \cdot d x-\int \frac{d}{d x} x \cdot \int \sec x \cdot \tan x \cdot d x \cdot d x\right) \\
& =x \cdot \tan x-\int(1) \cdot \tan x \cdot d x+x \cdot \sec x-\int(1) \cdot \sec x \cdot d x \\
& =x \cdot \tan x-\log (\sec x)+x \cdot \sec x-\log (\sec x+\tan x)+c \\
& =x \cdot(\sec x+\tan x)-\log (\sec x)-\log (\sec x+\tan x)+c \\
& \therefore \quad \int \frac{x}{1-\sin x} \cdot d x=x \cdot(\sec x+\tan x)-\log [(\sec x)(\sec x+\tan x)]+c \\
\end{aligned}
$
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