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Question: If \(\int{\dfrac{dx}{\sqrt{{{\sin }^{3}}x\cos x}}}=g\left( x \right)+c\) then \(g\left( x \right)\) ...

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

Explanation

Solution

We first convert the given integration to its required form of trigonometric expression. We use the differential form of tan⁡x\tan x and get the integral form using ∫xndx=xn+1n+1+c\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 ∫dxsin⁡3xcos⁡x\int{\dfrac{dx}{\sqrt{{{\sin }^{3}}x\cos x}}}.
We first multiply sec⁡2x{{\sec }^{2}}x to both the denominator and numerator of 1sin⁡3xcos⁡x\dfrac{1}{\sqrt{{{\sin }^{3}}x\cos x}}.
So, sec⁡2xsec⁡2xsin⁡3xcos⁡x=sec⁡2xsin⁡3xcos⁡3x=sec⁡2xtan⁡3/2x\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\tan x and get d(tan⁡x)=sec⁡2xdxd\left( \tan x \right)={{\sec }^{2}}xdx
We now rearrange the differential form and get

& \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{aligned}$$ Now we apply the differential formula $$\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1}+c$$ and get $$\begin{aligned} & \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{aligned}$$ 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.