How do you find the integral intsqrt(x^2-9)/x^3dx ?

1 Answer
Sep 1, 2014

By using the substitution x=3sec theta,
int{sqrt{x^2-9}}/{x^3}dx=1/6sec^{-1}(x/3)-{sqrt{x^2-9}}/{2x^2}+C.

Let x=3sec theta Rightarrow dx=3sec theta tan theta d theta.
int{sqrt{x^2-9}}/{x^3}dx=int{sqrt{(3sec theta)^2 -9}}/{(3sec theta)^3}3sec theta tan theta d theta,
which simplifies to
=1/3int{tan^2 theta}/{sec^2 theta}d theta
by using tan theta={sin theta}/{cos theta} and sec theta=1/{cos theta},
=1/3int sin^2 theta d theta
by using sin^2 theta ={1-cos(2theta)}/2,
1/6int[1-cos(2theta)]d theta=1/6[theta-sin(2theta)/2]+C
by using sin(2theta)=2sin theta cos theta,
=1/6(theta-sin theta cos theta)+C

Now, we need to rewrite our answer above in terms of x.
x=3sec theta Rightarrow x/3=sec theta Rightarrow theta=sec^{-1}(x/3)
Since sec theta=x/3, we can construct a right triangle with angle theta such that its hypotenuse is x, its adjacent is 3, and its opposite is sqrt{x^2-9}. So, we have
sin theta={sqrt{x^2-9}}/{x} and cos theta=3/x

Hence, by rewriting theta, sin theta, cos theta in terms of x,
int{sqrt{x^2-9}}/{x^3}dx=1/6sec^{-1}(x/3)-{sqrt{x^2-9}}/{2x^2}+C.