# Exam question - Integrals

• Feb 1st 2011, 07:53 AM
CMartins
Exam question - Integrals
The results of my calculus exam came and there is a question I had wrong.

Which is the integral of (sin(logx))/x?

My answer was -cos(logx) + c

I checked wolfram and the answer is the same, therefore that makes my answer correct right?
However, it says on the paper that this answer is wrong.
I did some research and realised that the derivative of -cos(logx) is not (sin(logx))/x, and maybe that's why it is wrong.
But how come I did not get the correct answer? Even wolfram says my answer is correct.

My resoltuion process was the following:

- turn (logx) into u;
- turn du into (1/x)dx -------> dx = xdu

- that gives me the integral of (xsen(u)/x);
- cutting the x's on either side gives me the integral of sen(u)
- resolving the integral gives me -cos(u)
- replacing the u gives me -cos(logx) and the add the constant +c

This method always never failed me, so what went wrong?

I have to go to college tomorrow to defend my answer, so I would like to get some replies fast... I need two points more on this exam to pass this subject.

Thank you.
• Feb 1st 2011, 07:55 AM
Ackbeet
Quote:

the derivative of -cos(logx) is not (sin(logx))/x
It's not?
• Feb 1st 2011, 08:00 AM
CMartins
It gives me sin(logx);
The original function multiplies that for 1/x
• Feb 1st 2011, 08:01 AM
Ackbeet
You're forgetting the chain rule.
• Feb 1st 2011, 08:02 AM
topsquark
I agree with both of you.

-Dan
• Feb 1st 2011, 08:08 AM
CMartins
@topsquark: I didn't talk to him yet. He just posted on the college website the results of each question. And apparently, I got that one wrong. I have to go there tomorrow to review my exam, that is why I need to prepare to defend my answer.

@Ackbeet: You're right! In that case my answer is correct right?
• Feb 1st 2011, 08:11 AM
Ackbeet
I would say your answer is correct. You should have been looking to use the chain rule to verify the integration, if the rule of integration used was a u substitution (they are inverses, just like by-parts is the inverse of the product rule).
• Feb 1st 2011, 08:13 AM
topsquark
I can't get anywhere with the idea but could this be a case of the arbitrary constant is different between the exam answer and CMartins's answer?

-Dan
• Feb 1st 2011, 08:17 AM
Ackbeet
Quote:

Originally Posted by topsquark
I can't get anywhere with the idea but could this be a case of the arbitrary constant is different between the exam answer and CMartins's answer?

-Dan

I don't know. It seems a fairly straight-forward problem to me. My guess is that the professor's key is wrong. There's a typo there or something.

By the way, I enjoyed the Dune quote in your sig.
• Feb 1st 2011, 08:18 AM
topsquark
It's just about the only thing from Dune that can actually be applied in the real world. It's a nice little piece for a simple meditative state when you are nervous.

-Dan
• Feb 1st 2011, 08:21 AM
topsquark
Quote:

Originally Posted by Ackbeet
I don't know. It seems a fairly straight-forward problem to me. My guess is that the professor's key is wrong. There's a typo there or something.

It probably is simply a typo on the answer key, but I have a vague memory of a problem involving logs where two answers were essentially the same, but the arbitrary constant made the two solutions look like they were different. But I don't remember the details.

-Dan
• Feb 1st 2011, 08:21 AM
CMartins
Thank you. I will talk to him tomorrow and resolve this.
• Feb 1st 2011, 08:24 AM
Ackbeet
You're very welcome for my contribution. Have fun!
• Feb 1st 2011, 09:38 AM
TheCoffeeMachine
Quote:

Originally Posted by topsquark
It probably is simply a typo on the answer key, but I have a vague memory of a problem involving logs where two answers were essentially the same, but the arbitrary constant made the two solutions look like they were different. But I don't remember the details.

Was it a case of the following? We have that $I = \int\frac{1}{ax+b}\;{dx} = \frac{1}{a}\ln\left|ax+b\right|+k_{1}.$ But also have
that $I = \int\frac{1}{ax+b}\;{dx} = \frac{1}{a}\int\frac{1}{x+\frac{b}{a}}\;{dx} = \frac{1}{a}\ln\left|x+\frac{b}{a}\right|+k_{2}.$ These are really the same because:
$\frac{1}{a}\ln\left(ax+b\right)+k_{1}$ $= \frac{1}{a}\ln\left|a\left(x+\frac{b}{a}\right)\ri ght|+k_{1}$ $=\frac{1}{a}\ln\left|x+\frac{b}{a}\right|+\frac{1} {a}\ln{a}+k_{1} = \frac{1}{a}\ln\left|x+\frac{b}{a}\right|+k_{2}.$
• Feb 1st 2011, 10:28 AM
topsquark
Yes, that might very well be the one I was thinking of. Thanks!

-Dan