Q68 / 160 · AP EAMCET 2014 · Mathematics

If $\int \frac{d x}{\sqrt{\sin ^3 x \cos x}}=g(x)+c$, then $g(x)$ is equal to

  • A
    $\frac{-2}{\sqrt{\cot x}}$
  • Correct answer
    B
    $\frac{-2}{\sqrt{\tan x}}$
  • C
    $\frac{2}{\sqrt{\cot x}}$
  • D
    $\frac{2}{\sqrt{\tan x}}$

Solution

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 $\quad \cot x=t$ $$ \begin{array}{rlrl} \Rightarrow-\operatorname{cosec}^2 x d x & =d t \\ \therefore \quad \int-\frac{1}{\sqrt{t}} d t & =-\frac{t^{1 / 2}}{1 / 2}+c \\ & =-2 \sqrt{\cot x}+c \\ & =-\frac{2}{\sqrt{\tan x}}+c \\ \therefore & g(x) & =-\frac{2}{\sqrt{\tan x}} \end{array} $$

Hence, the correct answer is option: 'B'.

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