# How do you find the antiderivative of #int (x^3sinx) dx#?

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To find the antiderivative of ( \int x^3 \sin(x) , dx ), you can use integration by parts method.

Let: ( u = x^3 ) ( dv = \sin(x) , dx )

Differentiate ( u ) to get ( du ), and integrate ( dv ) to get ( v ).

[ du = 3x^2 , dx ] [ v = -\cos(x) ]

Now, apply the integration by parts formula:

[ \int u , dv = uv - \int v , du ]

Substitute the values:

[ \int x^3 \sin(x) , dx = -x^3 \cos(x) - \int (-\cos(x))(3x^2 , dx) ]

Now integrate the remaining expression:

[ \int (-\cos(x))(3x^2 , dx) = 3 \int x^2 \cos(x) , dx ]

You can use integration by parts again for this expression.

Let: ( u = x^2 ) ( dv = \cos(x) , dx )

[ du = 2x , dx ] [ v = \sin(x) ]

Apply integration by parts:

[ \int x^2 \cos(x) , dx = x^2 \sin(x) - \int (\sin(x))(2x , dx) ]

[ = x^2 \sin(x) - 2 \int x \sin(x) , dx ]

You can use integration by parts one more time to evaluate ( \int x \sin(x) , dx ).

Let: ( u = x ) ( dv = \sin(x) , dx )

[ du = dx ] [ v = -\cos(x) ]

Apply integration by parts:

[ \int x \sin(x) , dx = -x \cos(x) - \int (-\cos(x))(dx) ]

[ = -x \cos(x) + \int \cos(x) , dx ]

[ = -x \cos(x) + \sin(x) + C ]

Substitute back into the previous expression:

[ \int x^3 \sin(x) , dx = -x^3 \cos(x) - 2(x^2 \sin(x) - 2(-x \cos(x) + \sin(x))) + C ]

[ = -x^3 \cos(x) - 2x^2 \sin(x) + 4x \cos(x) - 4\sin(x) + C ]

This is the antiderivative of ( \int x^3 \sin(x) , dx ).

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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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