By using the substitution x=3secθ,
∫√x2−9x3dx=16sec−1(x3)−√x2−92x2+C.
Let x=3secθ⇒dx=3secθtanθdθ.
∫√x2−9x3dx=∫√(3secθ)2−9(3secθ)33secθtanθdθ,
which simplifies to
=13∫tan2θsec2θdθ
by using tanθ=sinθcosθ and secθ=1cosθ,
=13∫sin2θdθ
by using sin2θ=1−cos(2θ)2,
16∫[1−cos(2θ)]dθ=16[θ−sin(2θ)2]+C
by using sin(2θ)=2sinθcosθ,
=16(θ−sinθcosθ)+C
Now, we need to rewrite our answer above in terms of x.
x=3secθ⇒x3=secθ⇒θ=sec−1(x3)
Since secθ=x3, we can construct a right triangle with angle θ such that its hypotenuse is x, its adjacent is 3, and its opposite is √x2−9. So, we have
sinθ=√x2−9x and cosθ=3x
Hence, by rewriting θ, sinθ, cosθ in terms of x,
∫√x2−9x3dx=16sec−1(x3)−√x2−92x2+C.
