Let x=2sin(u) implies dx = 2cos(u)dux=2sin(u)⇒dx=2cos(u)du
int (dx)/(x^2sqrt(4-x^2)) = int (2cos(u))/(4sin^2(u)sqrt(4 - 4sin^2(u)))du∫dxx2√4−x2=∫2cos(u)4sin2(u)√4−4sin2(u)du
= int (2cos(u))/(8sin^2(u)sqrt(1 - sin^2(u)))du=∫2cos(u)8sin2(u)√1−sin2(u)du
= int (2cos(u))/(8sin^2(u)sqrt(cos^2(u)))du=∫2cos(u)8sin2(u)√cos2(u)du
= int (2cos(u))/(8sin^2(u)cos(u))du=∫2cos(u)8sin2(u)cos(u)du
=1/4int (du)/(sin^2(u)) = 1/4int csc^2(u)du=14∫dusin2(u)=14∫csc2(u)du
You can look up that the integral of csc^2(x)csc2(x) is equal to -cot(x)−cot(x) but we shall derive this anyway.
int csc^2(x)dx = int (dx)/sin^2(x)∫csc2(x)dx=∫dxsin2(x)
Divide top and bottom by cos^2(x)cos2(x)
int (sec^2(x))/(tan^2(x))dx∫sec2(x)tan2(x)dx
Let z = tan(x) implies dz = sec^2(x)dxz=tan(x)⇒dz=sec2(x)dx
int (dz)/(z^2) = -1/z = -1/(tan(x)) = -cot(x)∫dzz2=−1z=−1tan(x)=−cot(x)
Hence:
1/4int csc^2(u)du = -1/4cot(u) + C14∫csc2(u)du=−14cot(u)+C
u = sin^(-1)(x/2)u=sin−1(x2)
therefore I = -1/4cot(sin^(-1)(x/2)) + C
This can look a little scary but it's actually quite simple, we just need to draw a triangle.
Consider y = sin^(-1)(a)
implies a = sin(y)
Sine is Opposite/Hypotenuse so we have triangle with opposite side a and Hypotenuse equal to 1. Can work out the other side with pythagoras' to be sqrt(1-a^2).
Now, the tangent function is opposite/adjacent, hence:
tan(y) = (a)/(sqrt(1-a^2))
cottheta = 1/(tantheta) implies cot(y) = (sqrt(1-a^2))/(a)
![enter image source here]()
So -1/4cot(sin^(-1)(x/2)) = -1/4(sqrt(1-(x/2)^2))/(x/2)
=-1/2(sqrt(1-(x^2/4)))/(x) = -1/4(sqrt(4-x^2))/x = -(sqrt(4-x^2))/(4x)
