# Trig substitution integration problem

• May 6th 2008, 10:50 AM
kdizzle711
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
• May 6th 2008, 12:09 PM
o_O
Before converting it directly into $\cot^{2} \theta$, look at it in terms of cos and sin:
$\int \frac{\cos^{2} \theta}{\sin^{2} \theta} d\theta = \int \frac{1-\sin^{2}\theta}{\sin^{2} \theta} d\theta = \int \left( \frac{1}{\sin^{2}\theta }- \frac{\sin^{2} \theta}{\sin^{2} \theta}\right)d\theta$

You can simplify it more and it'll be easier to integrate.
• May 6th 2008, 01:41 PM
kdizzle711
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?
• May 6th 2008, 02:23 PM
o_O
Recall that $\frac{d}{dx} \cot \theta = -\csc^{2} \theta$ ;)

So you can see what to do with your integral: $\int \left(\csc^{2} \theta - 1\right) d\theta$
• May 6th 2008, 04:00 PM
kdizzle711
Is it

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

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

-(sqrt(9-x^2)/x) - x
• May 6th 2008, 04:47 PM
o_O
Since you're integrating with respect to $\theta$, $\int 1 d\theta = \theta + C$ not $x + C$.

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

So using your initial sub, you have to find an expression for $\theta$ in terms of x.
• May 6th 2008, 05:21 PM
kdizzle711
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))

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