How do you integrate \[\int {\dfrac{{{x^3}}}{{\sqrt {1 - {x^2}} }}} \] using trigonometric substitution?

Hint: Here in this question, we have to integrate the given function. Substituting the terms by the trigonometry ratios and using or applying the trigonometry’s standard identities to the given function and then integrating the function. Hence, we obtain the required solution for the given question.

Complete step-by-step solution:
Consider the given function
\[\int {\dfrac{{{x^3}}}{{\sqrt {1 - {x^2}} }}} dx\]
Integrating the function by trigonometric substitution. Substitute \[x = \sin (u)\], so \[dx = \cos (u)du\]
On substituting x and dx
 \[ \Rightarrow \int {\dfrac{{{{\sin }^3}(u)}}{{\sqrt {1 - {{\sin }^2}(u)} }}} \cos (u)du\]----------(1)
By the standard trigonometry identity \[{\sin ^2}(u) + {\cos ^2}(u) = 1\]
This is written as \[{\cos ^2}(u) = 1 - {\sin ^2}(u)\]-----------(2)
Substituting the equation (2) in the equation (1), the function obtained as
\[ \Rightarrow \int {\dfrac{{{{\sin }^3}(u)}}{{\sqrt {{{\cos }^2}(u)} }}} \cos (u)du\]
The square root and the square is inverse to each other. Therefore, the square and square root will get cancels. The function is
\[ \Rightarrow \int {\dfrac{{{{\sin }^3}(u)}}{{\cos (u)}}} \cos (u)du\]
The cos (u) presents in the denominator and the denominator will get cancel. The function is written as
\[ \Rightarrow \int {{{\sin }^3}(u)} du\]
The \[{\sin ^3}(u)\]trigonometry ratio can be written as \[{\sin ^2}(u) \times \sin (u)\]. So the function is rewritten as
\[ \Rightarrow \int {{{\sin }^2}(u) \times \sin (u)} du\]
By the standard trigonometry identity \[{\sin ^2}(u) + {\cos ^2}(u) = 1\]
This is written as \[{\sin ^2}(u) = 1 - {\cos ^2}(u)\]
\[ \Rightarrow \int {(1 - {{\cos }^2}(u))\sin (u)} du\]
Substitute \[t = \cos (u)\], so \[dt = - \sin (u)du\]
On substituting
\[ \Rightarrow \int {(1 - {t^2})( - dt)} \]
\[ \Rightarrow - \int {(1 - {t^2})dt} \]
Applying the integration to each term we get
\[ \Rightarrow - \int {dt} + \int {{t^2}dt} \]
By the standard formulas on integration, we have \[\int {dt} = t\] and \[ \Rightarrow \int {{t^2}dt} = \dfrac{{{t^3}}}{3}\]. On substituting these integration formulas, we get
\[ \Rightarrow - t + \dfrac{{{t^3}}}{3} + c\]
Substitute \[t = \cos (u)\]
\[ \Rightarrow - \cos (u) + \dfrac{{{{\cos }^3}(u)}}{3} + c\]
As we know that \[x = \sin (u)\] so the value of u is given by \[u = {\sin ^{ - 1}}(x)\]. On substituting we get
\[ \Rightarrow - \cos ({\sin ^{ - 1}}(x)) + \dfrac{{{{\cos }^3}({{\sin }^{ - 1}}(x))}}{3} + c\]
This is rewritten as
\[ \Rightarrow - \dfrac{{\cos ({{\sin }^{ - 1}}(x))}}{3}\left( {3 - {{\cos }^2}({{\sin }^{ - 1}}(x)} \right) + c\]
As we know that \[\cos ({\sin ^{ - 1}}(x)) = \sqrt {1 - {x^2}} \], so the above function is written as
\[ \Rightarrow - \dfrac{{\sqrt {1 - {x^2}} }}{3}\left( {3 - {{\left( {\sqrt {1 - {x^2}} } \right)}^2}} \right) + c\]
The square root and the square are inverse to each other. Therefore, the square and square root will get canceled. The function is
\[ \Rightarrow - \dfrac{{\sqrt {1 - {x^2}} }}{3}\left( {3 - 1 + {x^2}} \right) + c\]
On simplification we get
\[ \Rightarrow - \dfrac{{\sqrt {1 - {x^2}} }}{3}\left( {2 + {x^2}} \right) + c\]
Hence we have integrated the given function and obtained the solution for the given question.

Note: By simplifying the question using the substitution we can integrate the given function easily. If we apply integration directly it may be complicated to solve further. So simplification is needed. We must know the differentiation formulas. The trigonometry standard identities are applied to solve this problem.