If $\int \dfrac{1}{\sqrt{\sin ^3 x \cos x}} d x=\dfrac{k}{\sqrt{\tan x}}+c$ then the value of $k$ is

Let $I = \int \dfrac{1}{\sqrt{\sin^{3} x \cos x}} dx$.

Rewrite the integrand as $I = \int \dfrac{1}{\sqrt{\sin^{2} x \cdot \sin x \cos x}} dx = \int \dfrac{1}{\sin x \sqrt{\sin x \cos x}} dx$.

Divide the numerator and denominator by $\cos^{2} x$ inside the square root or manipulate the expression as follows:

$I = \int \dfrac{1}{\sin x \sqrt{\dfrac{\sin x}{\cos x} \cdot \cos^{2} x}} dx = \int \dfrac{1}{\sin x \cos x \sqrt{\tan x}} dx$.

Multiply numerator and denominator by $\sec^{2} x$:

$I = \int \dfrac{\sec^{2} x}{\tan x \sqrt{\tan x}} dx = \int \dfrac{\sec^{2} x}{(\tan x)^{3/2}} dx$.

Substitute $u = \tan x$, then $du = \sec^{2} x dx$.

$I = \int u^{-3/2} du = \dfrac{u^{-1/2}}{-1/2} + c = -2 u^{-1/2} + c = -2 \dfrac{1}{\sqrt{\tan x}} + c$.

Comparing this with the given form $\dfrac{k}{\sqrt{\tan x}} + c$, we find $k = -2$.

Answer: $-2$