Using hyperbolic substitution to solve an integral

[Howard Fox]

Homework Statement

Homework Equations

So the question is asking to solve an integral and to use the answer of that integral to find an additional integral. With part a, I don't have much problem, but then I don't know how to apply the answer from it to part b. I know I should subsitute all the x's in part b with a trig function, right? But which one? [/B]

The Attempt at a Solution

This is my solution for part a, I am not sure how to work on part b

[/B]

 

[kuruman]

You are replacing $\sinh^2x$ with $1+\cosh^2x$. Check your algebra.

 

[Orodruin]

Also, the derivative of cosh(x) is sinh(x), not -sinh(x).

 

[Howard Fox]

Okay, I tried to correct the previous mistakes,

Hope this one is good? Thank you

 

[kuruman]

It is good. Look at part (a) and then look at part (b). What kind of hyperbolic substitution is worth trying in (b)?

 

[Howard Fox]

It is good. Look at part (a) and then look at part (b). What kind of hyperbolic substitution is worth trying in (b)?

I really don't know. I have trouble reconciling the two parts. I guess we could substitute x with asinh?

 

[kuruman]

I really don't know. I have trouble reconciling the two parts. I guess we could substitute x with asinh?

Excellent idea! What do you think $a$ ought to be? Hint: Look at the denominator.

 

[Howard Fox]

Excellent idea! What do you think $a$ ought to be? Hint: Look at the denominator.

It should be 3?

 

[Howard Fox]

Excellent idea! What do you think $a$ ought to be? Hint: Look at the denominator.

So the final integral should look something like this?

This does seem very complicated to solve though!

 

[kuruman]

It should be 3?

Yes.

So the final integral should look something like this?
View attachment 239808
This does seem very complicated to solve though!

That's because your algebra has been careless (again). Take the time to do it right. Then you will see that
(a) the denominator simplifies.
(b) you do not have a proper integral because you substituted $1$ for $dx$.
(c) the numerator is incorrect.

 

[Howard Fox]

Yes.

That's because your algebra has been careless (again). Take the time to do it right. Then you will see that
(a) the denominator simplifies.
(b) you do not have a proper integral because you substituted $1$ for $dx$.
(c) the numerator is incorrect.

Uhm, is this any better?

 

[kuruman]

Uhm, is this any better?

View attachment 239812

Much better. Now concentrate on simplifying the denominator. How could you get rid of the radical? How does one usually get rid of a radical?

 

[Howard Fox]

Much better. Now concentrate on simplifying the denominator. How could you get rid of the radical? How does one usually get rid of a radical?

Raising both denominator and numerator to the power of two?

 

[kuruman]

You would do that if the expression under the radical is not a perfect square. Certainly the $9$ in the denominator is a perfect square. What about the rest? Hint: Why did you choose $3$ for the substitution $x=3\sinh v$?

 

[Howard Fox]

You would do that if the expression under the radical is not a perfect square. Certainly the $9$ in the denominator is a perfect square. What about the rest? Hint: Why did you choose $3$ for the substitution $x=3\sinh v$?

/

Just take the radical of the values like this?

 

[kuruman]

/View attachment 239813
Just take the radical of the values like this?

Absolutely not ! Is $\sqrt{2^2+1^2}=2+1=3~$?

 

[Howard Fox]

Absolutely not ! Is $\sqrt{2^2+1^2}=2+1=3~$?

Okay, went back to my textbook. This should be the way to simplify the denominator, right?

 

[kuruman]

Right, except there is more carelessness. What is your integration variable?

 

[Howard Fox]

Right, except there is more carelessness. What is your integration variable?

It was dx but then we substituted it to du, no?