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$$
\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} d x=
$$
Options:
\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x)} d x=
$$
Solution:
2410 Upvotes
Verified Answer
The correct answer is:
$2 \log \left(\sin \frac{x}{2}\right)+C$
$\begin{aligned} & \text { Let } I=\int \sqrt{1+2 \cot x(\cot x+\operatorname{cosec} x} d x . \\ & =\int \sqrt{\operatorname{cosec}^2 x-\cot ^2 x+2 \cot ^2 x+2 \cot x \operatorname{cosec} x} d x \\ & =\int \sqrt{\operatorname{cosec}^2 x+\cot ^2 x+2 \cot x \operatorname{cosec} x} d y \\ & =\int(\operatorname{cosec} x+\cot x) d x \\ & =\log |\operatorname{cosec} x-\cot x|+\log |\sin x|+C \\ & =\log \left|\frac{1-\cos x}{\sin x} \times \sin x\right|+C\end{aligned}$
$\begin{aligned} & =\log \left|2 \sin ^2 \frac{x}{2}\right|+{ }^{\circ} C \\ & =\log 2+\log \sin ^2 \frac{x}{2}+C \\ & =2 \log \left|\sin \frac{x}{2}\right|+C[\because \log 2+C=C]
\end{aligned}$
$\begin{aligned} & =\log \left|2 \sin ^2 \frac{x}{2}\right|+{ }^{\circ} C \\ & =\log 2+\log \sin ^2 \frac{x}{2}+C \\ & =2 \log \left|\sin \frac{x}{2}\right|+C[\because \log 2+C=C]
\end{aligned}$
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