# Thread: Eek Urgent Integral Help!?!

1. ## Eek Urgent Integral Help!?!

Need help integrating:

integral(2cot(x/3)dx)

for my Calc I take home

so far, i've gotten as far as

2integral(cos[x/3]*(1/sin[x/3]))dx

or am i going in the wrong direction here?

2. Originally Posted by calcidiots101
Need help integrating:

integral(2cot(x/3)dx)

for my Calc I take home

so far, i've gotten as far as

2integral(cox[x/3]*(1/sin[x/3]))dx

or am i going in the wrong direction here?
In general,
$\int \cot t dt=\int \frac{\cos t}{\sin t}dt$
Let $u=\sin t$ then $u'=\cos t$
Thus, by the substitution rule,
$\ln |\sin t|+C$

Thus, when you have,
$\int 2 \cot (x/3) dx = 2\cdot 3 \ln |\sin (x/3)|+C$

3. see edit above, dno if it makes a difference...

also, could you explain that natural log substitution a bit further? i'm sort of fuzzy on it...

4. Originally Posted by calcidiots101
see edit above, dno if it makes a difference...

also, could you explain that natural log substitution a bit further? i'm sort of fuzzy on it...
Just like $u=\sin t$ and then just use the subsitution rule.

5. thanks for your great help so far...
but i don't really understand the substitution...

basically, i have:

u=sinx
du=cosxdx

so then that becomes:

integral([1/u]du)??

6. Originally Posted by calcidiots101
...anyone?
integral{(1/u) du} = ln(|u|) + c

now substitute back for u.

or
2integral(cos[x/3]*(1/sin[x/3]))dx

put v = sin[x/3]

dv = cos[x/3] * (1/3) dx

so that cos[x/3] * dx = 3 dv

2integral(cos[x/3]*(1/sin[x/3]))dx = 2 integral {3dv / v}

= 6 integral {dv / v}

= 6* ln(|v|)+ c

= 6* ln (| sin[x/3] |)+ c

some integration formulae are available at

calculus(integration formulae)

7. Originally Posted by calcidiots101
...anyone?
INT.[2cot(x/3)]dx -------------------(i)
= 2*INT.[cos(x/3) /sin(x/3)]dx -----------(ii)

You supposed u = sin(x/3)
So du = cos(x/3) *(1/3) dx
du = (1/3)cos(x/3)dx

Back to (ii),
2*INT.[cos(x/3) /sin(x/3)]dx
= 2*INT.[1 /sin(x/3)]*cos(x/3)dx
= 2*INT.[1/u]cos(x/3)dx -----------------(iii)

Now, cos(x/3)dx is not du.
The du is (1/3)cos(x/3)dx.
Meaning, the cos(x/3)dx is 3 times du.

a) Hence, so that you can use the cos(x/3)dx as "du", you have to divide the cos(x/3)dx by 3.
In doing that, you also have to multiply the [cos(x/3)dx]/3 by 3---so that you will not alter the (iii).
Meaning, you divide by 3, and also multiply by 3.
= 2*INT.[1/u]cos(x/3)dx *(3/3)
= (2*3)INT.[1/u]*(1/3)cos(x/3)dx
= 6*INT.[1/u]du
= 6*ln|u| +C
Since u = sin(x/3),