How do you integrate #int x^3 /sqrt(4 - 2x^2) dx# using trigonometric substitution?

Try the substitution: #x = sqrt2sin(u)#

This would mean #dx = sqrt2cos(u)dx#

Now put this into the integral and we get:

#int(2^(3/2)sin^3(u))/sqrt(4-4sin^2(u))sqrt2cos(u) du#

Tidying this up a little and using the trig -identity:
#sin^2(x)+cos^2(x) =1#, we get:

#int(4sin^3(u))/(2sqrt(1-sin^2(x)))cos(u)du#

#=2intsin^3u/(sqrt(cos^2(u)))cos(u)du=4intsin^3(u)du#

At this point we need another trig identity . The trig identity that we will use is: #sin^3x = 3/4sinx - 1/4sin(3x)#.

Putting that into the integral we get:

#2int3/4sinu-1/4sin(3u)dx#

Which can now be integrated to obtain:

#-3/2cos(u) + 1/6cos(3u)+C#

We obviously have the challenge now of reversing the substituting. Again using the trig identity near the start, this formula can be re - arranged to give:

#-3/2sqrt(1-sin^2(u))+1/6sqrt(1-sin^2(3u))+C#

At this point we can now rearrange the above trig identity to get:

#sin(3u) = 3sin(u) - 4sin ^3(u)#

We can now rewrite the above expression as:

#-3/2sqrt(1-sin^2(u))+1/6sqrt(1-(3sin(u)-4sin^3(u))^2)+C#

And finally we can reverse the substitution to obtain:

#-3/2sqrt(1-x^2/2)+1/6sqrt(1-(3/sqrt(2)x-sqrt2x^2)^2)+C#