Integration substiuition of new variable

[delsoo]

Homework Statement

for this question, my ans is pi/2 not pi/4 . can anybody please check where's the mistake?

Homework Equations

The Attempt at a Solution

 

[D H]

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

 

[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:
$dx = f(y)dy$

 

[delsoo]

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

it doesn't 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

 

[delsoo]

can anybody help please?

 

[D H]

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.

 

[delsoo]

i cheked thru my working again and again but stilll can't find my mistake... can you be more specific?

 

[D H]

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

 

[delsoo]

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

 

[D H]

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.

 

[delsoo]

i still can't understand how can it be wrong?

 

[delsoo]

how can 2yx not equal to 1?

 

[D H]

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 $\int\frac{-2\,dy}{\sqrt{1/4-y^2}}$.