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If $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$, then $g(x)$ is equal to
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Verified Answer
The correct answer is:
$\frac{-2}{\sqrt{\tan x}}$
Given, $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$
$$
\text { Now, } \begin{aligned}
\int \frac{d x}{\sqrt{\sin ^4 x \cot x}} & =\int \frac{d x}{\sin ^2 x \sqrt{\cot x}} \\
& =\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cot x}} d x
\end{aligned}
$$
Put $\quad \cot x=t$
$$
\begin{array}{rlrl}
\Rightarrow-\operatorname{cosec}^2 x d x & =d t \\
\therefore \quad \int-\frac{1}{\sqrt{t}} d t & =-\frac{t^{1 / 2}}{1 / 2}+c \\
& =-2 \sqrt{\cot x}+c \\
& =-\frac{2}{\sqrt{\tan x}}+c \\
\therefore & g(x) & =-\frac{2}{\sqrt{\tan x}}
\end{array}
$$
$$
\text { Now, } \begin{aligned}
\int \frac{d x}{\sqrt{\sin ^4 x \cot x}} & =\int \frac{d x}{\sin ^2 x \sqrt{\cot x}} \\
& =\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cot x}} d x
\end{aligned}
$$
Put $\quad \cot x=t$
$$
\begin{array}{rlrl}
\Rightarrow-\operatorname{cosec}^2 x d x & =d t \\
\therefore \quad \int-\frac{1}{\sqrt{t}} d t & =-\frac{t^{1 / 2}}{1 / 2}+c \\
& =-2 \sqrt{\cot x}+c \\
& =-\frac{2}{\sqrt{\tan x}}+c \\
\therefore & g(x) & =-\frac{2}{\sqrt{\tan x}}
\end{array}
$$
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