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:

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

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