Trig substitution integration problem, test in 1hr 30min

In summary, the conversation is about solving a trig substitution problem involving the integral of √(9-x²) / (x²). The solution involves using the substitution u = asinΘ and x = 3sinΘ, and then rewriting the equation in terms of Θ. The correct solution is found to be (x/3)ln(x/3)-ln(3/x+(√9-x²/x)), but there is some confusion about the steps involved in getting to this solution. It is mentioned that integrals can have different forms of answers, but they all have a common output.
  • #1
kdizzle711
27
0

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
 
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  • #2
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 [itex]\int \sqrt{9- x^2}/x^2 dx[/itex].

Yes, with [itex]x= 3sin(\theta)[/itex] [itex]\sqrt{9- x^2}= 3\sqrt{1- sin^2(\theta)}= 3cos(\theta)[/itex]. Now, what is x2 in terms of [itex]\theta[/itex]. You need to replace that too. And what is dx? You need to write that in terms of [itex]\theta[/itex].
 
  • #3
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))
 
  • #5
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
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
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 [Broken]

Integrals can have many forms of answers and this one is one of them but what they share is that they have common output.
 
Last edited by a moderator:
  • #8
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:

1. What is a trig substitution integration problem?

A trig substitution integration problem involves using trigonometric identities and substitution techniques to solve a complex integral. It is a common type of integration problem in calculus.

2. How do I approach a trig substitution integration problem?

The first step is to identify the appropriate substitution based on the form of the integral. Common substitutions include using the identities sin^2(x) + cos^2(x) = 1 or tan^2(x) + 1 = sec^2(x). Once the substitution is made, the integral can be evaluated using standard integration techniques.

3. What are some common mistakes to avoid when solving a trig substitution integration problem?

Some common mistakes include forgetting to substitute back in the original variable at the end, not using the correct trigonometric identity, and missing a negative sign in front of the square root term.

4. How should I prepare for a trig substitution integration problem on a test?

Practice solving a variety of problems and make sure you have a solid understanding of the different trigonometric identities and their derivatives. It can also be helpful to create a cheat sheet with the most commonly used substitutions and identities.

5. How much time should I spend on a trig substitution integration problem during a test?

It is recommended to spend no more than 10-15 minutes on a single problem during a test. If you are struggling to make progress, it may be best to move on and come back to it later. Time management is key in successfully completing a test with a limited time frame.

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