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

Question

Scarlett Allison · Accepted Answer

#int4x^3sin(x^4)dx=-cos(x^4)+C#

We need to know that the antiderivative of #sin(u)# is #-cos(u)#, that is:

#intsin(u)du=-cos(u)+C#

We have the problem:

#int4x^3sin(x^4)dx#

Use the substitution #u=x^4#. This implies that #du=4x^3dx#. This, luckily, is already in the integrand:

#int4x^3sin(x^4)dx=intsin(x^4)(4x^3dx)=intsin(u)du#

Which is an integral we can work with:

#int4x^3sin(x^4)dx=-cos(u)+C#

Returning to our original variable:

#int4x^3sin(x^4)dx=-cos(x^4)+C#

Christopher Agresta · Answer

undefined 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 $.

Charles Anctil · Answer

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

1. Let $ u = x^4 $, then $ du = 4x^3 \, dx $.
2. Substitute $ u = x^4 $ and $ du = 4x^3 \, dx $ into the integral.
3. Rewrite the integral in terms of $ u $, resulting in $ \int \sin(u) \, du $.
4. Integrate $ \sin(u) $ with respect to $ u $ to get $ -\cos(u) + C $.
5. 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.