Question

If we have an integral as $\int{\dfrac{\sqrt{1-{{x}^{2}}}}{{{x}^{4}}}}dx=A\left( x \right).{{\left( \sqrt{1-{{x}^{2}}} \right)}^{m}}+C$ for a suitable integer m and a function A(x), where C is a constant of integration ${{\left( A\left( x \right) \right)}^{m}}$ equals:
(a) $\dfrac{-1}{3{{x}^{3}}}$
(b) $\dfrac{-1}{27{{x}^{9}}}$
(c) $\dfrac{1}{9{{x}^{4}}}$
(d) $\dfrac{1}{27{{x}^{6}}}$

Answer 1

To solve this question, we need to find the value of A(x) and the value of m and to find the given values, we need to find solve the given integral. To solve this integral we will employ the substitution method. Once, we are able to solve the integral, we shall find the value in the form $A\left( x \right).{{\left( \sqrt{1-{{x}^{2}}} \right)}^{m}}$. Thus, by comparison, we will find the value of A(x) and m and once we have these values of A(x) and m, we can easily find the value of ${{\left( A\left( x \right) \right)}^{m}}$

Complete step-by-step solution:
The integral given to us is as follows:
$\Rightarrow \int{\dfrac{\sqrt{1-{{x}^{2}}}}{{{x}^{4}}}}dx$
We can take one of the x from ${{x}^{4}}$ of the denominator and take it inside the root. The integral will change as follows:
$\Rightarrow \int{\dfrac{1}{{{x}^{3}}}\left[ \sqrt{\dfrac{1}{{{x}^{2}}}\left( 1-{{x}^{2}} \right)} \right]}dx$
We will solve the parenthesis inside the square root.
$\Rightarrow \int{\dfrac{1}{{{x}^{3}}}\left[ \sqrt{\dfrac{1}{{{x}^{2}}}-1} \right]}dx$
Now, let t = $\dfrac{1}{{{x}^{2}}}-1$.
We will now differentiate both sides to find the substitution for dx.
$\Rightarrow$ dt = $\dfrac{-2}{{{x}^{3}}}$dx
$\Rightarrow$ dx = $\dfrac{-{{x}^{3}}}{2}$dt
Now, we will make the substitution, the integral changes as follows:
$\Rightarrow \int{\dfrac{1}{{{x}^{3}}}\left[ \sqrt{t} \right]}\left( \dfrac{-{{x}^{3}}}{2} \right)dt$
We can now divide out ${{x}^{3}}$ from the numerator and the denominator. We shall also take 2 from the denominator and the negative sign from the numerator outside the integral sign as they will not affect the integral.
$\Rightarrow \dfrac{-1}{2}\int{\left[ \sqrt{t} \right]}dt$
Now, this is known integral and thus we shall now execute the integration.
$\begin{aligned} & \Rightarrow \dfrac{-1}{2}\int{\left[ \sqrt{t} \right]}dt=\dfrac{-1}{2}\left[ \dfrac{{{t}^{\dfrac{1}{2}+1}}}{\dfrac{1}{2}+1} \right]+C \\\ & \Rightarrow \dfrac{-1}{2}\int{\left[ \sqrt{t} \right]}dt=\dfrac{-1}{2}\left[ \dfrac{{{t}^{\dfrac{3}{2}}}}{\dfrac{3}{2}} \right]+C \\\ & \Rightarrow \dfrac{-1}{2}\int{\left[ \sqrt{t} \right]}dt=-\dfrac{{{t}^{\dfrac{3}{2}}}}{3}+C \\\ \end{aligned}$
We can now reverse the substitution and write the equation in terms of x.
$\Rightarrow \int{\dfrac{\sqrt{1-{{x}^{2}}}}{{{x}^{4}}}}dx=-\dfrac{{{\left( \dfrac{1}{{{x}^{2}}}-1 \right)}^{\dfrac{3}{2}}}}{3}$
Now, we will simplify the right hand side to get it in the form $A\left( x \right).{{\left( \sqrt{1-{{x}^{2}}} \right)}^{m}}$.
$\begin{aligned} & \Rightarrow -\dfrac{{{\left( \dfrac{1}{{{x}^{2}}}-1 \right)}^{\dfrac{3}{2}}}}{3}=\dfrac{-{{\left( \sqrt{1-{{x}^{2}}} \right)}^{3}}}{3{{x}^{3}}} \\\ & \Rightarrow \int{\dfrac{\sqrt{1-{{x}^{2}}}}{{{x}^{4}}}}dx=\dfrac{-1}{3{{x}^{3}}}.{{\left( \sqrt{1-{{x}^{2}}} \right)}^{3}}+C \\\ \end{aligned}$
On comparing the right hand side, we get that A(x) = $\dfrac{-1}{3{{x}^{3}}}$ and m = 3.
Therefore, ${{\left( A\left( x \right) \right)}^{m}}={{\left( -\dfrac{1}{3{{x}^{3}}} \right)}^{3}}=-\dfrac{1}{27{{x}^{9}}}$
Hence, option (b) is the correct option.

Note: The given integral is a complex integral and substation is the best method to solve it. Students can also go for integration by parts but it will further complicate the integral. Students are advised to properly select the substitute so that extra terms get divided out.