By using the substitution x=3sec thetax=3secθ,
int{sqrt{x^2-9}}/{x^3}dx=1/6sec^{-1}(x/3)-{sqrt{x^2-9}}/{2x^2}+C∫√x2−9x3dx=16sec−1(x3)−√x2−92x2+C.
Let x=3sec theta Rightarrow dx=3sec theta tan theta d thetax=3secθ⇒dx=3secθtanθdθ.
int{sqrt{x^2-9}}/{x^3}dx=int{sqrt{(3sec theta)^2 -9}}/{(3sec theta)^3}3sec theta tan theta d theta∫√x2−9x3dx=∫√(3secθ)2−9(3secθ)33secθtanθdθ,
which simplifies to
=1/3int{tan^2 theta}/{sec^2 theta}d theta=13∫tan2θsec2θdθ
by using tan theta={sin theta}/{cos theta}tanθ=sinθcosθ and sec theta=1/{cos theta}secθ=1cosθ,
=1/3int sin^2 theta d theta=13∫sin2θdθ
by using sin^2 theta ={1-cos(2theta)}/2sin2θ=1−cos(2θ)2,
1/6int[1-cos(2theta)]d theta=1/6[theta-sin(2theta)/2]+C16∫[1−cos(2θ)]dθ=16[θ−sin(2θ)2]+C
by using sin(2theta)=2sin theta cos thetasin(2θ)=2sinθcosθ,
=1/6(theta-sin theta cos theta)+C=16(θ−sinθcosθ)+C
Now, we need to rewrite our answer above in terms of xx.
x=3sec theta Rightarrow x/3=sec theta Rightarrow theta=sec^{-1}(x/3)x=3secθ⇒x3=secθ⇒θ=sec−1(x3)
Since sec theta=x/3secθ=x3, we can construct a right triangle with angle thetaθ such that its hypotenuse is xx, its adjacent is 33, and its opposite is sqrt{x^2-9}√x2−9. So, we have
sin theta={sqrt{x^2-9}}/{x}sinθ=√x2−9x and cos theta=3/xcosθ=3x
Hence, by rewriting thetaθ, sin thetasinθ, cos thetacosθ in terms of xx,
int{sqrt{x^2-9}}/{x^3}dx=1/6sec^{-1}(x/3)-{sqrt{x^2-9}}/{2x^2}+C∫√x2−9x3dx=16sec−1(x3)−√x2−92x2+C.
