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

We have to integrate by parts, or apply the reverse product rule as I like to call it, in order to integrate this function.

I won't go into too much detail here, but you can get more assistance by watching the video from Khan Academy.

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To find the integral of (x \sin^2(x)), you can use integration by parts. Let (u = x) and (dv = \sin^2(x)dx). Then, (du = dx) and (v = \frac{x}{2} - \frac{\sin(2x)}{4}). Applying the integration by parts formula, (\int u , dv = uv - \int v , du), you'll get the result:

[ \int x \sin^2(x) , dx = -\frac{x^2}{2} \sin^2(x) + \frac{x}{2} \sin(2x) - \frac{1}{4} \cos(2x) + 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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