# How do you integrate #int x^2sin^2x#?

After using tabular integration for second integral.

Thus,

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To integrate ( \int x^2 \sin^2(x) ), you can use integration by parts twice.

Let ( u = x^2 ) and ( dv = \sin^2(x) dx ). Then, ( du = 2x , dx ) and ( v = \frac{1}{2}(x - \sin(2x)) ).

Using the integration by parts formula ( \int u , dv = uv - \int v , du ), you get:

[ \int x^2 \sin^2(x) , dx = \frac{1}{2}x^2\sin(2x) - \int x \left(x - \frac{1}{2}\sin(2x)\right) , dx ]

Integrating the remaining integral by parts again:

Let ( u = x ) and ( dv = \left(x - \frac{1}{2}\sin(2x)\right) dx ). Then, ( du = dx ) and ( v = \frac{1}{2}\left(x^2 - \frac{1}{4}\cos(2x)\right) ).

Applying integration by parts again, you get:

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

Simplify the expression:

[ \int x^2 \sin^2(x) , dx = \frac{1}{4}x^2\sin(2x) + \frac{1}{8}\cos(2x) - \frac{1}{4}x\sin(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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