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$\int \frac{\sqrt{\cot x}}{\sin x \cos x} d x=-f(x)+c \Rightarrow f(x)$
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2655 Upvotes
Verified Answer
The correct answer is:
$2 \sqrt{\cot x}$
Given that,
$$
\begin{aligned}
& \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x=-f(x)+c \\
& \Rightarrow \quad \int \frac{\sqrt{\cot x}}{\cot x \sin ^2 x} d x=-f(x)+c \\
& \Rightarrow \quad \int \frac{1}{\sqrt{\cot x \sin ^2 x}} d x=-f(x)+c \\
& \Rightarrow \quad-2 \sqrt{\cot x}+c=-f(x)+c \\
& \Rightarrow \quad f(x)=2 \sqrt{\cot x} \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \int \frac{\sqrt{\cot x}}{\sin x \cos x} d x=-f(x)+c \\
& \Rightarrow \quad \int \frac{\sqrt{\cot x}}{\cot x \sin ^2 x} d x=-f(x)+c \\
& \Rightarrow \quad \int \frac{1}{\sqrt{\cot x \sin ^2 x}} d x=-f(x)+c \\
& \Rightarrow \quad-2 \sqrt{\cot x}+c=-f(x)+c \\
& \Rightarrow \quad f(x)=2 \sqrt{\cot x} \\
&
\end{aligned}
$$
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