integral of x^3//(1-x^2)
i got trig substitution to be me to the integral of sine thetea cubed. after that I would take one of the sines and convert. I would take sine to the second power. I changed this to 1-cos^2. Now I am stuck.
I have the integral of (1-cos^2x)/(sin^x). any input?

Question

integral of x^3//(1-x^2)
i got trig substitution to be me to the integral of sine thetea cubed. after that I would take one of the sines and convert. I would take sine to the second power. I changed this to 1-cos^2. Now I am stuck.
I have the integral of (1-cos^2x)/(sin^x). any input?

hartnn · Answer

is your original Q this : 
$\int \dfrac{x^3}{\sqrt{1-x^2}}dx$

hartnn · Answer

if you put x= sin t there, you'd get 
$\int \sin^3 t \: dt$
and you can use the sin 3x formula.
sin 3x = 3sin x -4 sin^3 x 
isolate sin^3 x from here and substitute in your integral.

raden · Answer

alternative : 
we can solve by u subs ..
let u=1-x^2 ---> du = -2x dx or x dx = -du/2
x^2 = 1-u
so, ur integral can be :
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