# How do you find the integral of #x^3 * sin( x^2 ) dx#?

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To find the integral of (x^3 \cdot \sin(x^2) , dx), you can use the technique of substitution. Let (u = x^2), then (du = 2x , dx). Rearranging, we have (dx = \frac{du}{2x}). Substituting these expressions into the integral, we get:

[ \int x^3 \sin(x^2) , dx = \int \frac{1}{2} u \sin(u) , du ]

This integral can be solved using integration by parts. Let (dv = \sin(u) , du) and (u = \frac{1}{2}u). Then (v = -\frac{1}{2} \cos(u)). Applying the integration by parts formula, we have:

[ \int \frac{1}{2}u \sin(u) , du = -\frac{1}{2}u \cos(u) - \int -\frac{1}{2}\cos(u) , du ]

[ = -\frac{1}{2}x^2 \cos(x^2) + \frac{1}{2} \int \cos(u) , du ]

[ = -\frac{1}{2}x^2 \cos(x^2) + \frac{1}{2} \sin(u) + C ]

[ = -\frac{1}{2}x^2 \cos(x^2) + \frac{1}{2} \sin(x^2) + C ]

Where (C) is the constant of integration.

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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.

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.

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.

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