#int 1/(x^2 sqrt(25-x^2)) dx =?#
IN PARTICULAR (this case)
Note that #25-x^2 ge 0 <=> x in [-5,5]#.
In this case, the useful substitution is
#5 sin theta=x#
and this is invertible, because for #theta in [-pi/2,pi/2]# the function #x(theta)=5 sin theta# is invertible with inverse function #theta(x) = arcsin(x/5)#, where #x in [-5,5]#. So this is a valid substitution.
Let's compute the differential term
#5 cos theta * d theta =dx#
and turn the integral to #theta#, substituting the expressions found:
#int 1/(x^2 sqrt(25-x^2)) dx=int 1/(25 sin^2 theta sqrt(25-25 sin^2 theta)) 5 cos theta * d theta = int (cos theta * d theta)/(5 sin^2 theta * 5 sqrt(1- sin^2 theta))=1/25 int (cos theta * d theta)/(sin^2 theta cos theta) = - 1/25 int -(d theta)/(sin^2 theta)=-1/25 cot theta + C#
Now back to #x#:
#int 1/(x^2 sqrt(25-x^2)) dx=-1/25 cot theta + C=-1/25 cot(arcsin(x/5)) + C#
From trigonometry #cot(arcsin(x/5))=(sqrt(1-(x/5)^2))/(x/5)#.
In fact, you can think of a right triangle with hypotenuse of length #1# and one leg of length #x/5# for #0 < x < 5#. By Pythagorean Theorem the length of the other leg is #sqrt(1-(x/5)^2)#. If we call #alpha# the angle between this #sqrt(1-(x/5)^2)# leg and the hypotenuse, where #0 < alpha < pi/2#, then by definition of sine we have that #sin alpha = (x/5)/1=x/5#, which means that #alpha=arcsin(x/5)#. Moreover, #cot(alpha)=sqrt(1-(x/5)^2)/(x/5)#.
So if we substitute #alpha# in this equality we get the result for #0 < x < 5#. The result can be extended to #x in [-5,5]# thanks to the symmetry of the functions involved or by re-thinking this argument on the unit circle.
In the end, the result is:
#int 1/(x^2 sqrt(25-x^2)) dx=-1/25 (sqrt(1-(x/5)^2))/(x/5)+C=-1/25 (sqrt(25-x^2))/x+C#
IN GENERAL
When facing expressions like #sqrt(a-bx^2)#, where #a,b > 0#, you can take advantage of trigonometric substitutions. Remember that the existence of the radical is determined by the non-negativity of #a-bx^2#. We have that
#a-bx^2 ge 0 <=> -sqrt(a/b) le x le sqrt(a/b)#
This will be important to guarantee that the substitution works with invertible functions.
Let's work on the radical and rewrite it in the following way:
#sqrt(a-bx^2)=sqrt(a[1-b/a x^2])=sqrt(a[1-(sqrt(b/a) x)^2])=sqrt(a)sqrt(1-(sqrt(b/a) x)^2)#
Now, if #sin theta = sqrt(b/a) x#, we can take advantage of the Pythagorean Trigonometric Identity
#cos^2 theta + sin^2 theta =1 <=> cos^2 theta = 1-sin^2 theta=1-(sqrt(b/a) x)^2#
This means that
#sqrt(a-bx^2)=sqrt(a)sqrt(1-(sqrt(b/a) x)^2)=sqrt(a) sqrt(cos^2 theta)#
Notice that for #theta \in [-pi/2,pi/2]# the function #sin theta# is invertible and #sin theta in [-1,1]#. Moreover, for #x in [-sqrt(a/b),sqrt(a/b)]#, the function #sqrt(b/a) x# is invertible and #sqrt(b/a) x in [-1,1]#. So the substitution holds, because we're working with invertible functions mapped to the same interval.
The cosine function is non-negative for #-pi/2 le theta le pi/2#, so we get that
#sqrt(a-bx^2)=sqrt(a) cos theta#
So the substitution #sin theta = sqrt(b/a) x# always "removes" radicals of this kind.
