How do you integrate #((4-x^2)^(1/2)) / x^2#?

1 Answer
Aug 24, 2015

Keeping trigonometric substitution in mind, the numerator is of the form:

#sqrt(a^2 - x^2)#

which resembles #sqrt(1-sin^2x)#. Thus, let:

#x = asintheta# where #a = 2#
#=> x = 2sintheta#
#x^2 = a^2sin^2theta = 4sin^2theta#
#sqrt(4-x^2) = sqrt(2^2-2^2sin^2theta) = 2costheta#
#dx = 2costhetad theta#

With this substitution, we get:

#= int (cancel(2)costheta)/(cancel(4)sin^2theta) cancel(2)costhetad theta#

#= int cot^2thetad theta#

#= int csc^2theta - 1d theta#

since #1 + cot^2theta = csc^2theta#, just like how #1 + tan^2theta = sec^2theta#.

The derivative of #cottheta# is #-csc^2theta#, so with some negative-sign manipulation, we can get this into a more simplified form:

#= -int 1-csc^2thetad theta#

#= -(int d theta - intcsc^2thetad theta)#

#= -(int d theta + int-csc^2thetad theta)#

#= -int d theta - int-csc^2thetad theta#

#= -theta - cottheta#

With #x = 2sintheta# and #sqrt(4-x^2) = 2costheta#:

#=> theta = arcsin(x/2)#
#=> cottheta = costheta/sintheta = (sqrt(4-x^2)/cancel(2))(cancel(2)/x)#
#= sqrt(4-x^2)/x#

So the final answer is:

#= color(blue)(-arcsin(x/2) - sqrt(4-x^2)/x + C)#