# Trig substitution integration problem, test in 1hr 30min

1. May 6, 2008

### kdizzle711

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

√(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

√9cos²Θ=3cosΘ/x^2

where do I go from here, I tried getting help at Math forum, but they were complete tools. Plus I have had a good experience with this site in the past

2. May 6, 2008

### HallsofIvy

Staff Emeritus
You do understand, don't you that many of the people who post here, also post at other forums?

Since you say this is an integral, √(9-x²) / (x²) is NOT the problem statement: it has to be something like $\int \sqrt{9- x^2}/x^2 dx$.

Yes, with $x= 3sin(\theta)$ $\sqrt{9- x^2}= 3\sqrt{1- sin^2(\theta)}= 3cos(\theta)$. Now, what is x2 in terms of $\theta$. You need to replace that too. And what is dx? You need to write that in terms of $\theta$.

3. May 6, 2008

### kdizzle711

I still think I am making a mistake somewhere

I did
x²=9sin²Θ

so------- 3cosΘ/9sin²Θ= (1/3)cotΘcscΘ

int(1/3)cotΘcscΘ

=(x/3)ln(sinu)-ln(cscu+cotu)
using the triangle i found
(x/3)ln(x/3)-ln(3/x+(√9-x²/x))

4. May 6, 2008

5. May 6, 2008

### kdizzle711

I forgot to take into acount the dx, but when I do i get

(3cosΘ/9sin²Θ)*(3cosΘ)=9cos²Θ/9sin²Θ= cot²

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

I am stuck

6. May 6, 2008

### racer

x = 3siny

dx = 3cosy dy

(9-9siny^2)^1/2. 3cosy dy/9siny^2

(9-9siny^2) = 9cosy^2 because cosy^2 + sin ^2 = 1, multiply both sides by 9.

it becomes

3cosy . 3 cos y dy / 9 siny^2

9cosy^2 dy / 9 siny^2

it becomes cosy^2 dy / sin^2 which is coty^2 dy

but 1 + coty^2 = cscy^2. coty^2 = cscy^2 - 1

now integrate this cscy^2 - 1

you get -cot(y) - y.

I am not so sure about my solution.

7. May 6, 2008

### racer

Integrals can have many forms of answers and this one is one of them but what they share is that they have common output.

8. May 6, 2008

### BrendanH

What Racer has done/said is correct; in this case, his solution is right, and integrals can be of different form. I agree after having made an error the first time :S

Last edited: May 6, 2008