Question

How do you use substitution to integrate \[{x^2}\left( {\sin {x^3}} \right)\] ?

Answer 1

Hint: Here the hint is in the given question only. We have to use the substitution the only question is what to substitute? Now we will substitute \[u = {x^3}\] because the derivation of the substitution helps us to solve the integration very easily. This is one of the methods that helps to find the integration of a complicated problem.
Formula used:
 \[\dfrac{d}{{dx}}{x^n} = \dfrac{{{x^{n + 1}}}}{{n + 1}}\]
 \[\int {\sin x} = - \cos x\]

Complete step by step solution:
Given that \[\int {{x^2}\left( {\sin {x^3}} \right)dx} \]
Now we will substitute \[u = {x^3}\]
Taking the derivative with respect to x using the formula mentioned above we get,
 \[du = 3{x^2}dx\]
Then \[dx = \dfrac{{du}}{{3{x^2}}}\]
Substituting the values in the integration above,
 \[ = \int {{x^2}\sin u} \dfrac{{du}}{{3{x^2}}}\]
Cancelling the \[{x^2}\] term,
 \[ = \dfrac{1}{3}\int {\sin udu} \]
We know that \[\int {\sin x} = - \cos x\]
 \[ = - \dfrac{{\cos u}}{3} + C\]
Resubstituting the value of u we get,
 \[ = - \dfrac{{\cos {x^3}}}{3} + C\]
This is the answer.
So, the correct answer is “$- \dfrac{{\cos {x^3}}}{3} + C$”.

Note: Here note that the substitution is the method so mentioned in the question. If that method is not mentioned specifically then we can use integration by parts with the sequence of functions like ILATE that means Inverse,logarithmic, algebraic, trigonometric and last the exponential. This is the sequence to be followed.