Question

How do you evaluate the integral of $\int {\dfrac{x}{{{{(1 - {x^4})}^{\dfrac{1}{2}}}}}} dx$ ?

Answer 1

We use some concepts of integration and some methods of integration to solve this problem. And mainly we use substitution methods in this problem.We will also learn some standard formulas and results that are present in integration and also differentiation. We use the formula $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$

Complete step by step answer:
In mathematics, integration is a reverse process of differentiation.
Here we need to integrate the term $\dfrac{x}{{{{(1 - {x^4})}^{\dfrac{1}{2}}}}}$ and let the integral be $I$
So, $I = \int {\dfrac{x}{{\sqrt {1 - {x^4}} }}} dx$
Now, let us rewrite this as
$I = \int {\dfrac{x}{{\sqrt {1 - {{({x^2})}^2}} }}} dx$ -----(1)
Here, $\dfrac{d}{{dx}}{x^2} = 2x$ , it means that the term ${x^2}$ and its derivative also exist in the question.

So, take ${x^2} = t$ which on differentiating on both sides gives us $2xdx = dt$
And from here, we can conclude that, $xdx = \dfrac{1}{2}dt$
Now, substitute these values in (1)
$\Rightarrow I = \int {\dfrac{{\dfrac{1}{2}dt}}{{\sqrt {1 - {t^2}} }}}$
$\Rightarrow I = \dfrac{1}{2}\int {\dfrac{{dt}}{{\sqrt {1 - {t^2}} }}}$ (As constant can be taken out of an integral)
We all know that, $\dfrac{d}{{dx}}{\sin ^{ - 1}}x = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
So, that implies $\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x} + c$

So, from this formula, we can conclude that,
$I = \dfrac{1}{2}\int {\dfrac{{dt}}{{\sqrt {1 - {t^2}} }}} = \dfrac{1}{2}{\sin ^{ - 1}}t + c$
Now, substituting back the original value of $t$ , we get,
$I = \dfrac{1}{2}{\sin ^{ - 1}}({x^2}) + c$
So, this is the required solution.
$\therefore \int {\dfrac{x}{{{{(1 - {x^4})}^{\dfrac{1}{2}}}}}} dx = \dfrac{1}{2}{\sin ^{ - 1}}({x^2}) + c$

Let us differentiate our answer to check whether we get our initial question or not.
So, $\dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\sin }^{ - 1}}({x^2}) + c} \right) = \dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\sin }^{ - 1}}({x^2})} \right) + \dfrac{d}{{dx}}c$
Constant comes out of the term and differentiation of a constant is 0.
So, $\dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\sin }^{ - 1}}({x^2}) + c} \right) = \dfrac{1}{2}\dfrac{d}{{dx}}\left( {{{\sin }^{ - 1}}({x^2})} \right)$
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\sin }^{ - 1}}({x^2}) + c} \right) = \dfrac{1}{2}\left( {\dfrac{1}{{\sqrt {1 - {{({x^2})}^2}} }}\dfrac{d}{{dx}}({x^2})} \right)$
Since $\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)$
So, we get it as
$\dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\sin }^{ - 1}}({x^2}) + c} \right) = \dfrac{1}{2}\left( {\dfrac{{2x}}{{\sqrt {1 - {x^4}} }}} \right)$
$\therefore \dfrac{d}{{dx}}\left( {\dfrac{1}{2}{{\sin }^{ - 1}}({x^2}) + c} \right) = \dfrac{x}{{\sqrt {1 - {x^4}} }}$

Therefore, the value of integral $\int {\dfrac{x}{{{{(1 - {x^4})}^{\dfrac{1}{2}}}}}} dx$ is
$\dfrac{1}{2}{\sin ^{ - 1}}({x^2}) + c$.

Note: Always remember to add an arbitrary constant to your answer that you got after integrating. Also remember some of the standard results like $\int {\dfrac{1}{{\sqrt {1 - {x^2}} }}dx = {{\sin }^{ - 1}}x + c}$ and $\int {\dfrac{{ - 1}}{{\sqrt {1 - {x^2}} }}dx = {{\cos }^{ - 1}}x + c}$. In a function, if there is a pair of terms and its derivative, then take that term as substitution. Because, when you take that term as substitution, its derivative also vanishes from the question and our problem will get easier.