∫5x
^{3}√(1-x
^{2})
dx
So I'm still a beginner with integrating by substituting.
What I know is that you use 'u' from one of the thing in the equation to substitute.
Though, I'm not entirely sure where 'u' should be coming from.
So I did
u = (1-x^{2})
And since it has a cube root, I did this:
1/3
du =
^{3}√(1-x
^{2})
dx
So I got: ∫1/3 u
^{1/3} du
4/3 u
^{4/3} + C -> 4/3 (1-x
^{2})
^{3} + C
But I'm not sure if this is entirely correct after:
Someone else showed me how to do the steps, which was this:
∫5x
^{3}√(1-x
^{2})
dx = -5/2∫(1-x
^{2})
^{1/3}(-2x)
dx = -5/2 * ((1-x
^{2})
^{4/3})/(4/3) + C =
(-15/8) * (1-x^{2})^{4/3} + C
This was the correct answer in the book.
But I'm a bit confused on how the person got the -5/2, and the (-2x)
Can someone explain to me that part?
Thanks!