Trig substitution integration problem, test in 1hr 30min

[kdizzle711]

Homework Statement

√(9-x²) / (x²)

Homework Equations

Just trig substitution

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

 

[HallsofIvy]

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.

 

[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))

 

[kdizzle711]

The right answer is here, but I'm not sure which steps i am messing up


http://integrals.wolfram.com/index.jsp?expr=sqrt(9-x^2)/(x^2)

 

[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

 

[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.

 

[racer]

The right answer is here, but I'm not sure which steps i am messing up


http://integrals.wolfram.com/index.j...%2F%28x%5E2%29

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

 

[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