Integration substiuition of new variable

1. May 6, 2014

delsoo

1. The problem statement, all variables and given/known data
for this question, my ans is pi/2 not pi/4 . can anybody please check where's the mistake?

2. Relevant equations

3. The attempt at a solution

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2. May 6, 2014

D H

Staff Emeritus
You didn't transform the integration limits, for one thing. I didn't look beyond that to see if you made other mistakes.

3. May 6, 2014

BiGyElLoWhAt

you also didn't solve properly for dx.
$x = \frac{1}{2y} ∴ \frac{dx}{dy} = ?$
just take the derivative of both sides with respect to y, and you'll end up with something like:
$dx = f(y)dy$

4. May 7, 2014

delsoo

it doesnt matter actually... eventually i change the x to y and then to tetha, lastly i change my limit for my tetha, what's wrong with my working . i ended up getting pi/2

5. May 7, 2014

6. May 7, 2014

D H

Staff Emeritus
Your factor of two error arises on the very first line, where you went from $\int \frac{dx}{x\sqrt{x^2-1}}$ to $-2\int\frac{dy}{\sqrt{1/4-y^2}}$ under the substitution $x=\frac 1 {2y}$. That factor of two is incorrect. You should have obtained $\int\frac{-dy}{\sqrt{1/4-y^2}}$ for that first step.

7. May 7, 2014

delsoo

i cheked thru my working again and again but stilll cant find my mistake... can you be more specific?

8. May 7, 2014

D H

Staff Emeritus
Show how you derived that very first step. That very first step is the source of your error.

9. May 7, 2014

delsoo

since x=(1/2y) , then i get 2xy =1 ... anything wrong?

10. May 7, 2014

D H

Staff Emeritus
Yes. Your first step is wrong. Show exactly how you went from $\int \frac{dx}{x\sqrt{x^2-1}}$ to $\int\frac{-2 \,dy}{\sqrt{1/4-y^2}}$ with that substitution.

11. May 7, 2014

delsoo

i still cant understand how can it be wrong?

12. May 7, 2014

delsoo

how can 2yx not equal to 1?

13. May 7, 2014

D H

Staff Emeritus
Somewhere along the way you did something wrong, but since you didn't show your intermediate steps, there's no knowing where you went wrong. That factor of two in your very first step is incorrect. You should have found that $\int \frac{dx}{x\sqrt{x^2-1}}$ with the substitution $2xy=1$ results in $\int\frac{-dy}{\sqrt{1/4-y^2}}$ rather than $\int\frac{-2\,dy}{\sqrt{1/4-y^2}}$.