# Thread: Trig substitution integration problem

1. ## Trig substitution integration problem

1. The problem statement, all variables and given/known data

integral√(9-x²) / (x²)

2. Relevant equations

Just trig substitution

3. The attempt at a solution

Ok, for trig sub I did

u=asinΘ
x=3sinΘ
9-x²=9-9sin²=9(1-sin²Θ)
so putting it into the equation
x²=9sin²Θ
dx=3cosΘ
√9cos²Θ=3cosΘ/x^2
(3cosΘ/9sin²Θ)*(3cosΘ)=9cos²Θ/9sin²Θ= cot²

but the integral of cot is ln(sinU) or it can be -ln(cscu)

I am stuck

2. Before converting it directly into $\displaystyle \cot^{2} \theta$, look at it in terms of cos and sin:
$\displaystyle \int \frac{\cos^{2} \theta}{\sin^{2} \theta} d\theta = \int \frac{1-\sin^{2}\theta}{\sin^{2} \theta} d\theta = \int \left( \frac{1}{\sin^{2}\theta }- \frac{\sin^{2} \theta}{\sin^{2} \theta}\right)d\theta$

You can simplify it more and it'll be easier to integrate.

3. So would it end up

int(csc²-1)

cot²=csc²-1

= int(cot²)

= ln(sinu)*ln(sinu)

= ln(x/3) * ln(x/3)

That still does not look right. What am I missing?

4. Recall that $\displaystyle \frac{d}{dx} \cot \theta = -\csc^{2} \theta$

So you can see what to do with your integral: $\displaystyle \int \left(\csc^{2} \theta - 1\right) d\theta$

5. Is it

$\displaystyle \int \left(\csc^{2} \theta - 1\right) d\theta$

$\displaystyle \left(-cot \theta - x\right) d\theta$

-(sqrt(9-x^2)/x) - x

6. Since you're integrating with respect to $\displaystyle \theta$, $\displaystyle \int 1 d\theta = \theta + C$ not $\displaystyle x + C$.

So your antiderivative is: $\displaystyle -\cot \theta - \theta + C$.

So using your initial sub, you have to find an expression for $\displaystyle \theta$ in terms of x.

7. So

x=3sin(theta)
(x/3)=sin(theta)
sin^-1(x/3)=theta

So do I just plug that in to
-cot(sin^-1(x/3))-(sin^-1(x/3))

8. Yep you can simplify it more though, specifically the $\displaystyle \cot \left[\sin^{-1} \left(\frac{x}{3}\right) \right]$ part. Recalling that $\displaystyle \cot \theta = \frac{\cos \theta}{\sin \theta}$
$\displaystyle \cot \left[\sin^{-1} \left(\frac{x}{3}\right) \right] = \frac{\cos \left[\sin^{-1} \left(\frac{x}{3}\right) \right]}{\sin \left[\sin^{-1} \left(\frac{x}{3}\right) \right]}$
Can you simplify it more? Your solution is correct. Just doesn't look that neat with that $\displaystyle \sin^{-1}\left(\frac{x}{3}\right)$ inside the $\displaystyle \cot x$.