Question

If $\int{\dfrac{dx}{\sqrt{{{\sin }^{3}}x\cos x}}}=g\left( x \right)+c$ then $g\left( x \right)$ is
A. $\dfrac{-2}{\sqrt{\cot x}}$
B. $\dfrac{-2}{\sqrt{\tan x}}$
C. $\dfrac{2}{\sqrt{\cot x}}$
D. $\dfrac{2}{\sqrt{\tan x}}$

Answer 1

Hint: We first convert the given integration to its required form of trigonometric expression. We use the differential form of $\tan x$ and get the integral form using \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c\]. We take the function leaving the constant.

Complete answer:
We complete the integration of $\int{\dfrac{dx}{\sqrt{{{\sin }^{3}}x\cos x}}}$.
We first multiply ${{\sec }^{2}}x$ to both the denominator and numerator of $\dfrac{1}{\sqrt{{{\sin }^{3}}x\cos x}}$.
So, $\dfrac{{{\sec }^{2}}x}{{{\sec }^{2}}x\sqrt{{{\sin }^{3}}x\cos x}}=\dfrac{{{\sec }^{2}}x}{\sqrt{\dfrac{{{\sin }^{3}}x}{{{\cos }^{3}}x}}}=\dfrac{{{\sec }^{2}}x}{{{\tan }^{{}^{3}/{}_{2}}}x}$.
We take the differential form of $\tan x$ and get $d\left( \tan x \right)={{\sec }^{2}}xdx$
We now rearrange the differential form and get
\[\begin{align}
  & \int{\dfrac{dx}{\sqrt{{{\sin }^{3}}x\cos x}}} \\
 & =\int{\dfrac{{{\sec }^{2}}xdx}{{{\tan }^{{}^{3}/{}_{2}}}x}} \\
 & =\int{\dfrac{d\left( \tan x \right)}{{{\left( \tan x \right)}^{{}^{3}/{}_{2}}}}} \\
 & =\int{{{\left( \tan x \right)}^{{}^{-3}/{}_{2}}}d\left( \tan x \right)} \\
\end{align}\]
Now we apply the differential formula \[\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c\] and get
\[\begin{align}
  & \int{{{\left( \tan x \right)}^{{}^{-3}/{}_{2}}}d\left( \tan x \right)} \\
 & =\dfrac{{{\left( \tan x \right)}^{{}^{-1}/{}_{2}}}}{{}^{-1}/{}_{2}}+c \\
 & =\dfrac{-2}{\sqrt{\tan x}}+c \\
\end{align}\]
We get $\int{\dfrac{dx}{\sqrt{{{\sin }^{3}}x\cos x}}}=g\left( x \right)+c=\dfrac{-2}{\sqrt{\tan x}}+c$. This gives $g\left( x \right)=\dfrac{-2}{\sqrt{\tan x}}$.
Therefore, the correct option is B.

Note:
We need to be careful about taking the differential form. We could also change the variable by taking $u=\tan x$. The differential also changes with the variable change. We have to replace the variable with the old form at the end to get the solution.