What is the solution to this natural logarithm homework problem?

[heyman123]

i have a problem please help

http://img373.imageshack.us/img373/3052/index1bd9.gif


Who can solve this for me?!

 

[berkeman]

We don't give out answers here on the PF. You need to show us your own work before we can offer tutorial help.

What are your thoughts on how to approach this problem?

 

[heyman123]

i can do it but i need the primitive of sin x and that's my problem, the rest of the problem i can solve , i just only can't manage to discover de primitive of sin x

 

[Defennder]

Is that a logarithm of base 10 or the natural logarithm?

 

[Defennder]

The indefinite integral doesn't appear to have an elementary solution. So I'm thinking there must be something special about the definite integral.

 

[Dick]

This is basically a trick. There is no elementary primitive. Here's a clue. Change the range of integration to 0 to pi/2 and call the integral I. Then you want -2*I. Now observe the integral of log(cos(x)) is also I. That's the clue. Add integral log(sin(x)) and log(cos(x)) and use a rule of logarithms and a trig identity and a u-substitution. Now you got an equation with a bunch of I's in it. Can you solve for I?

 

[physixguru]

Integrate by partial fraction.

 

[rock.freak667]

This is basically a trick. There is no elementary primitive. Here's a clue. Change the range of integration to 0 to pi/2 and call the integral I. Then you want -2*I. Now observe the integral of log(cos(x)) is also I. That's the clue. Add integral log(sin(x)) and log(cos(x)) and use a rule of logarithms and a trig identity and a u-substitution. Now you got an equation with a bunch of I's in it. Can you solve for I?

how come if you change from pi to pi/2, the integral is doubled? I get that if the integrand was just sinx but isn't lnsinx something entirely different?

 

[Dick]

ln(sin(x)) is just as symmetric as sin(x). The integral of it from 0->pi/2 is half the integral of it from 0->pi.

 

[rock.freak667]

oh ok then...I thought the graphs were very different.

 

[Dick]

They are very different. But they are still symmetric around x=pi/2. Did you solve the problem? It's really not that hard if you put your mind to it and know the secret hint. I only knew it because I've seen this problem before.

 

[rock.freak667]

They are very different. But they are still symmetric around x=pi/2. Did you solve the problem? It's really not that hard if you put your mind to it and know the secret hint. I only knew it because I've seen this problem before.

I only solved it based on your hints but I didn't know it was symmetric at pi/2. But how you knew to change the limits of integration beats me

 

[Dick]

Because cos(x) is negative between pi/2 and pi. So log(cos(x)) isn't defined. It just seemed neater to restrict the range rather than put an absolute value in. That's all.

 

[HallsofIvy]

i can do it but i need the primitive of sin x and that's my problem, the rest of the problem i can solve , i just only can't manage to discover de primitive of sin x

Surely, you didn't get to where you are being expect to solve problems like this without learning that the derivative of cos x is -sin x??

 

[dirk_mec1]

Surely, you didn't get to where you are being expect to solve problems like this without learning that the derivative of cos x is -sin x??

We already answered this below!

https://www.physicsforums.com/showthread.php?t=240644