Thread: Finding the integrand

1. Finding the integrand

$
\int_ \frac{\pi}{6}^ \frac{\pi}{2} cot(x)dx
$

I ended up switching the cot to the inverse of tangent. Then I put in sines/cosine for tangent. I then did a trig substitution.

u=sin(x)
du=cos(x)dx

However, I end up with this:

$ln1 - ln(1/2)$

ln1 is zero so I'm left with -ln(1/2).

The answer choices (this is mult.choice)

A. ln(1/2)
B. ln(2)
C. -ln(2-(sq.root of 3))
D. ln((sq.root of 3) -1)
E. None of the above

My answer key says A is the right answer, but I have NEGATIVE ln(1/2). I'm not sure where I went wrong. ??

2. Just switch cot(x) = cos(x)/sin(x) Let u =sin(x)

you end up with ln|sin(x)| an eventually ln1 - ln(1/2) = ln1 - ln1 +ln(2)

as ln(1/2) = ln1 = ln(2) So B is the right answer

3. Can you clarify this part?

- ln(1/2) = - ln1 +ln(2)

EDIT: btw you're right, my teacher said B is the answer

4. Your solution is correct! (Answer choice "A" is not correct.)

Using some log rules:

$\ln\left(\frac{a}{b}\right) = \ln(a) - \ln(b)$

You can find that

$-\ln(1/2) = \ln(2)$