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

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Answer 1 · 2018-04-20T01:19:13

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

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#

Answer 2 · 2018-04-20T08:00:38

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

#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#