please explain how to integrate (x^3)(1-x^4)^7 dx...?

Answer 1

There are two methods, Method 1 you could simplify (1-X^4)^7 then multiply through by X^3, but... you dont want to do that, especially on an exam.

Method 2 is substitution.

Lets make up an equation here that we can substitute into our original equation...

Say U=(1-X^4) so we have (X^3)U^7dx, understand?

Now lets take the derivative of U.... so dU= -4x^3dx, and solving for dx = dU/(-4x^3)

Lets plug that into our equation for dx.

we now have (X^3)*U^7dU/(-4x^3)

Canceling the X^3 were left with (-1/4)*U^7dU

Now you can integrate that right!

(1/8)*(-1/4)*U^8

further simplified is

(-1/32)U^8

Now just plug back in U=(1-X^4) and we get

(-1/32)*(1-X^4)^8

That should be the right answer!

Answer 2

the derivative of (1-x^4) = -4x^3 if you state u = (1-x^4)
then du/dx = -4x^3 and x^3 =- ((du/dx )/3)

so ;we can rewrite integrate 1/3 u^7 du

integration of u^7 = 1/8 u^8

the result is - 1/24 ((1-x^4)^8)

Answer 3

The easist way to iscalntegrate i think is to multiply out first.

remembering pascal triangle, then integrate.