Question

when integrating x*csc^2 (3x) i get two answers, one with ln and the other with log. Since cot(u) equals cos(u)/sin(u), I should get the same answer going either way, right?

Answer 1

\[\int\limits xcsc^2(3x)dx = -\frac{ 1 }{ 3 } xcot(3x) +\frac{ 1 }{ 3 } \int\limits \cot(3x)dx\]

Answer 2

What do you mean by "going either way?"

Answer 3

since \[\int\limits \cot(u) du = \ln \left| \sin(u) \right| + C\]
u = 3x
du = 3dx
1/3 du = dx

\[\int\limits xcsc^2 (3x) dx = \frac{ 1 }{ 9 } \ln \left| \sin(3x) \right|-\frac{ 1 }{ 3 }\cot(3x)x + C\]

Answer 4

however, if I use the trig identity cot(u) = cos(u)/sin(u) then I end up with log instead of ln. these answers aren't equivalent. Am I making a mistake here?

Answer 5

Well,
\[\int \frac{dt}{t}=\ln|t|+C\]
or more generally,
\[\int\frac{g'(t)}{g(t)}\,dt=\ln|g(t)|+C\]
The base on the logarithm will always be \(e\).

Answer 6

Depending on the literature, you might see the logarithm's notation as \(\ln x=\log x\), where \(e\) is the implicit base (instead of 10, which you might find elsewhere).

If that's the case, then \(\log x=\log_e x\not=\log_{10}x\).

Answer 7

I assumed that if I saw log it meant base 10... When I plugged this into wolfram alpha it gave me log instead of ln.

Wolfram alpha used substitution to solve the integral of cos(u)/sin(u) and found the integral of 1/s, thus giving the log.

I should always get the same answer regardless of which method I used to solve, right? Obviously the answer I got and the answer alpha gave aren't equivalent

Answer 8

I'm studying for a test so I want to make sure I use the correct method to solve a problem like this.

Answer 9

W|A uses the \(\log\) notation for \(\ln\). If it were any other base, it would explicitly say so.

Answer 10

Ooh, awesome. That answers my question then. Thanks!!!

Answer 11

yw