How do you find the integral of int4x^3 sin(x^4) dx?

2 Answers
Apr 20, 2018

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

Explanation:

The following substitution will do:

u=x^4

du=4x^3dx

int4x^3sin(x^4)dx=intsin(x^4)4x^3dx

So, we see this is a valid substitution, as du shows up in the integral.

We get

intsinudu=-cosu+C

Rewriting in terms of x yields

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

Apr 20, 2018

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

Explanation:

int4x^3sin(x^4)dx

we note that teh function in front of sin(x^4)
is the x^4 differentiated, so we can do this by inspection

d/(dx)(cosu)=-sinu

so we suspect the integral is of the form

cos(x^4)

d/(dx)(cos(x^4))=-4x^3sin(x^4)

:.int4x^3sin(x^4)dx=-cos(x^4)+c