How do I evaluate int1/[x^2(sqrt(25-x^2))] dx ?

1 Answer

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.