Quadratic Expressions: Advanced techniques of Integration

Question

$$\int \frac{x}{\sqrt{5+12x-9x^2}}\,dx$$

After two steps I arrive at $\displaystyle{ \int \frac{x}{\sqrt{9-(3x-2)^2}}}\,dx$

Using trigonometric substitution, we have a triangle with a cosine of $\theta$ of $\displaystyle{\sqrt{9-(3x-2)^2}\over 3}$ and a $\sin(\theta$) of $(3x-2)\over (3)$.

The final two steps of solving give

$\displaystyle {\frac{-1}{3}\cos(\theta) + \frac{2}{9}\theta +C}$

Therefore

-$\displaystyle{\sqrt{9-(3x-2)^2}\over 9}$

Wolfram alpha says that there should be a positive sign, i.e. -$\displaystyle{\sqrt{9+(3x-2)^2}\over 9}$

How can that be possible given the value for cosine of theta?

Answer 1

$$3I=\int\frac{3xdx}{\sqrt{3^2-(3x-2)^2}}$$

Setting $3x-2=3\sin\theta\implies dx=\cos\theta\ d\theta$

and $3x=2+3\sin\theta$

$$3I=\int\frac{\cos\theta(2+3\sin\theta)}{3\text{sign}(\cos\theta)\cos\theta}d\theta$$

$$\implies9I=\text{sign}(\cos\theta)\int(2+3\sin\theta)d\theta$$

$$=\text{sign}(\cos\theta)(2\theta-3\cos\theta+K)$$

Now, $\theta=\arcsin\dfrac{3x-2}3$ and $\cos^2\theta=1-\left(\dfrac{3x-2}3\right)^2=\dfrac{5+12x-9x^2}9$

Form the principal value definition, $-\dfrac\pi2\le\theta\le\dfrac\pi2\implies\cos\theta\ge0$

Answer 2

setting $$\sqrt{5+12x-9x^2}=xt+\sqrt{5}$$ we get $$x=\frac{12-2t\sqrt{5}}{t^2+9}$$ and we get $$dx=\frac{2(-9\sqrt{5}-12t+\sqrt{5}t^2)}{(9+t^2)^2}dt$$ and our integral is rational.