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$\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} d x=$
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Verified Answer
The correct answer is:
$2 \log \left|\sin \frac{x}{2}\right|+c$
We have,
$$
\begin{aligned}
I & =\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} d x \\
I & =\int \sqrt{1+\cot ^2 x+2 \operatorname{cosec} x \cot x+\cot ^2 x} d x \\
I & =\int \sqrt{(\operatorname{cosec} x+\cot x)^2} d x \\
I & =\int(\operatorname{cosec} x+\cot x) d x \\
I & =\log (\operatorname{cosec} x-\cot x)+\log \sin x+c \\
I & =\log (\operatorname{cosec} x-\cot x) \sin x+c \\
I & =\log (1-\cos x)+c \\
\Rightarrow I & =\log \left(2 \sin ^2 \frac{x}{2}\right)+c \Rightarrow I=2 \log \left|\sin \frac{x}{2}\right|+c
\end{aligned}
$$
$$
\begin{aligned}
I & =\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} d x \\
I & =\int \sqrt{1+\cot ^2 x+2 \operatorname{cosec} x \cot x+\cot ^2 x} d x \\
I & =\int \sqrt{(\operatorname{cosec} x+\cot x)^2} d x \\
I & =\int(\operatorname{cosec} x+\cot x) d x \\
I & =\log (\operatorname{cosec} x-\cot x)+\log \sin x+c \\
I & =\log (\operatorname{cosec} x-\cot x) \sin x+c \\
I & =\log (1-\cos x)+c \\
\Rightarrow I & =\log \left(2 \sin ^2 \frac{x}{2}\right)+c \Rightarrow I=2 \log \left|\sin \frac{x}{2}\right|+c
\end{aligned}
$$
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