# Tricky integral

1. Oct 29, 2008

### jbowers9

1. The problem statement, all variables and given/known data
I recently tried to do the following integral:
an = ∫√(2/a) sin(n∏x/a) cosh(x) dx
x=0 to x=a

2. Relevant equations
an = ∫√(2/a) sin(βx) cosh(x) dx
β = n∏/a
sin(βx) = ½i(eiβx – e-iβx)
cosh(x) = ½(ex + e-x)

3. The attempt at a solution

an = ¼ i √(2/a)∫ (eiβx – e-iβx) (ex + e-x)

after all is said and done, I get;

an = √(2/a)[(a2sin(n∏)sinh(a) – acos(n∏)cosh(a) + n∏a)/(n22 + a2)]

The text, “Quantum Mechanics Demystified”, however, gets;

an = √(2/a)[a(n∏cos(n∏)cosh(a) + sin(n∏)sinh(a))/( n22 + a2)]

Which is correct? And why?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited: Oct 30, 2008
2. Oct 30, 2008

### Staff: Mentor

Your approach is the one I would take, so here is what I would do:

Check your work to see if you can find any errors.
Take the derivative of your result. Do you get the integrand?
Take the derivative of the book's result. Do you get the integrand?
If the answers to both questions are yes, the two antiderivatives are equal or differ by a constant.
If one answer is yes and the other is no, the result from the "yes" answer is almost surely correct and the other is incorrect. It's even possible that the answer in the book is wrong.

3. Oct 30, 2008

### gabbagabbahey

If you've written the Integral correctly, then your solution is closer than the one from the text; it should have an aSin(....) term instead of an a2Sin(...)....Of course, if n is an integer then the sin (n*pi) term is zero and cos(n*pi)=(-1)n.

So, are you sure you are evaluating the correct integral?

4. Oct 30, 2008

### gabbagabbahey

Usually these are good strategies for checking a solution, but in this case the integral is a definite integral, so differentiating the textbook's solution will naturally give zero.

5. Oct 30, 2008

### jbowers9

I made an error transcribing the above and corrected the cosh(x) term. I've redone it 3 times and still get my results.