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

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Answer 1 · 2017-03-19T19:03:11

#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#