How do you integrate #int 1/(x^2sqrt(4-x^2))# by trigonometric substitution?

1 Answer
Jul 26, 2016

#int (dx)/(x^2sqrt(4-x^2)) = -(sqrt(4-x^2))/(4x) + C#

Explanation:

Let #x=2sin(u) implies dx = 2cos(u)du#

#int (dx)/(x^2sqrt(4-x^2)) = int (2cos(u))/(4sin^2(u)sqrt(4 - 4sin^2(u)))du#

#= int (2cos(u))/(8sin^2(u)sqrt(1 - sin^2(u)))du#

#= int (2cos(u))/(8sin^2(u)sqrt(cos^2(u)))du#

#= int (2cos(u))/(8sin^2(u)cos(u))du#

#=1/4int (du)/(sin^2(u)) = 1/4int csc^2(u)du#

You can look up that the integral of #csc^2(x)# is equal to #-cot(x)# but we shall derive this anyway.

#int csc^2(x)dx = int (dx)/sin^2(x)#

Divide top and bottom by #cos^2(x)#

#int (sec^2(x))/(tan^2(x))dx#

Let #z = tan(x) implies dz = sec^2(x)dx#

#int (dz)/(z^2) = -1/z = -1/(tan(x)) = -cot(x)#

Hence:

#1/4int csc^2(u)du = -1/4cot(u) + C#

#u = sin^(-1)(x/2)#

#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)#

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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)#