Evaluate the Integration:
\[\int_{0}^{{{\left( \dfrac{\pi }{2} \right)}^{\dfrac{1}{3}}}}{{{x}^{2}}sin{{x}^{3}}dx}\]

Hint: Integral contains function and its derivative then we use substitution method of integration to reduce the integral into standard form, so use method of substitution to solve it easily.

Complete step-by-step answer:
Let us consider the given integral as \[\int_{0}^{{{\left( \dfrac{\pi }{2} \right)}^{\dfrac{1}{3}}}}{{{x}^{2}}sin{{x}^{3}}dx}\ldots \ldots \ldots \left( 1.1 \right)\].
It consists of two functions ${{x}^{2}}$ and $sin{{x}^{3}}$. We know that the derivative of ${{x}^{3}}$ function consists of another function i.e. ${{x}^{2}}$. So, we can use the method of integration by substitution. In this method to reduce the function into standard form, let us consider \[{{x}^{3}}=t\] (here t is any arbitrary variable)…(1.2)
Differentiating both side w.r.t x and we know that $\dfrac{d\left( {{x}^{n}} \right)}{dx}$ = \[n{{x}^{n-1}}\]. Then we can write
${x}^{3} = t$
$\dfrac{d{{x}^{3}}}{dx}=\dfrac{dt}{dx}$
$3{{x}^{2}}=\dfrac{dt}{dx}$
Now, we can write it as \[3{{x}^{2}}dx=dt......\left( 1.3 \right)\]
Now, we have to change the limit with respect to the ‘t’ variable as we are going to change the expression in terms of ‘t’ using equation (1.2).
When,
$x=0$, then we have $t=0$
$x={{\left( \dfrac{\pi }{2} \right)}^{\dfrac{1}{3}}}$, then put the value of x in ${x}^{3} = t$ so we get $t=\dfrac{\pi }{2}$
Substituting the equation from (1.3) in equation (1.1). Then, we get
$\int\limits_{0}^{\dfrac{\pi }{2}}{\sin t.\dfrac{dt}{3}}$
$=\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{sin\text{ }t\text{ }dt}$
We know that $\int{sin\ t\ dt}=\ -cos\ t$, so,
$=\dfrac{1}{3}\int_{0}^{\dfrac{\pi }{2}}{sin\ t\ dt}$
$=-\dfrac{1}{3}[cos\ t]_{0}^{\dfrac{\pi }{2}}$
$=-\dfrac{1}{3}\left[ cos\left( \dfrac{\pi }{2} \right)-\ cos0 \right]$
We know, $cos\ \dfrac{\pi }{2}=0$ and $cos\ 0=1$, so we get
$\begin{align}
  & =-\dfrac{1}{3}(0-1) \\
 & =\dfrac{1}{3} \\
\end{align}$
So, therefore we have $\int_{0}^{{{\left( \dfrac{\pi }{2} \right)}^{\dfrac{1}{3}}}}{{{x}^{2}}sin{{x}^{3}}dx}\ =\dfrac{1}{3}$.

Note: The following points are to be noted while solving this type of question.
1) Use an appropriate method, such as we used the substitution method to get the answer easily.
2) In equation 1.2 differentiate the equation w.r.t x only, not w.r.t to t.
3) Don’t forget to change the limits in terms of ‘t’.