Eek Urgent Integral Help!?!

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January 29th 2007, 07:27 PM
calcidiots101

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?
January 29th 2007, 07:43 PM
ThePerfectHacker

Quote:

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,
$\int 2 \cot (x/3) dx = 2\cdot 3 \ln |\sin (x/3)|+C$
January 29th 2007, 07:51 PM
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...
January 29th 2007, 07:58 PM
ThePerfectHacker

Quote:

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.
January 29th 2007, 08:07 PM
calcidiots101

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)??
January 29th 2007, 09:48 PM
qpmathelp

Quote:

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)
January 29th 2007, 10:22 PM
ticbol

Quote:

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),
= 6*ln|sin(x/3)| +C ------------answer.

b) Or, if you want to wrack your brain more, :), multiply the du by 3, so that it will match the cos(x/3)dx.
Now, multiplying du by 3 will not affect, will not alter, the (iii). You are only substituting 3du for cos(x/3)dx into (iii), and 3du = cos(x/3)dx anyway.
So,
2*INT.[1/u]cos(x/3)dx
= 2*INT.[1/u](3du)
= (2*3)INT.[1/u]du
etc.....
January 30th 2007, 10:59 AM
calcidiots101

thanks so much for the help!