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How do you use substitution to integrate #(x^2)(sinx^3) #?

Answer 1

William Abdalla

#int (x^2)(sinx^3) dx = -1/3cosx^3 +C#

#int (x^2)(sinx^3) dx#

Let #u = x^3# so that #du = 3x^2 dx#

Out integral becomes:

#1/3 int (sinx^3) 3x^2dx = 1/3 int sin u du#

# = 1/3 (-cosu) +C#

# = -1/3cosx^3 +C#

Check the answer by differentiating:

#d/dx(-1/3cosx^3 +C) = -1/3* -sin(x^3) * 3x^2#

# = sin(x^3)* x^2#

Looks good, so that's it.

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Answer 2

Axel Alvarez

To integrate (x^2 \sin(x^3)) using substitution, you can use the following steps:

Let (u = x^3).
Calculate (du/dx), which is (3x^2).
Rearrange the equation to solve for (dx), yielding (dx = du/(3x^2)).
Substitute (u) and (dx) into the integral, resulting in (\int x^2 \sin(u) \cdot \frac{1}{3x^2} du).
Simplify the expression to (\frac{1}{3} \int \sin(u) du).
Integrate (\sin(u)) with respect to (u) to get (-\frac{1}{3} \cos(u) + C).
Finally, substitute back (u = x^3) to get the final answer: (-\frac{1}{3} \cos(x^3) + C), where (C) is the constant of integration.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.