# How do you integrate #int 4x^3sinx^4 dx#?

We have the problem:

Which is an integral we can work with:

Returning to our original variable:

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To integrate ( \int 4x^3 \sin(x^4) , dx ), you can use the substitution method. Let ( u = x^4 ), then ( du = 4x^3 , dx ). Substituting these into the integral, it becomes ( \int \sin(u) , du ), which is (-\cos(u) + C), where ( C ) is the constant of integration. Finally, resubstitute ( u = x^4 ) to get the final result: ( -\cos(x^4) + C ).

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To integrate ( \int 4x^3 \sin(x^4) , dx ), you can use a substitution method:

- Let ( u = x^4 ), then ( du = 4x^3 , dx ).
- Substitute ( u = x^4 ) and ( du = 4x^3 , dx ) into the integral.
- Rewrite the integral in terms of ( u ), resulting in ( \int \sin(u) , du ).
- Integrate ( \sin(u) ) with respect to ( u ) to get ( -\cos(u) + C ).
- Substitute back ( u = x^4 ) to obtain the final result.

The integral becomes:

[ \int \sin(u) , du = -\cos(u) + C ]

Substituting ( u = x^4 ) back:

[ \int 4x^3 \sin(x^4) , dx = -\cos(x^4) + C ]

So, the integral of ( 4x^3 \sin(x^4) ) with respect to ( x ) is ( -\cos(x^4) + 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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