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?

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- January 29th 2007, 08:27 PMcalcidiots101Eek 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, 08:43 PMThePerfectHacker
- January 29th 2007, 08:51 PMcalcidiots101
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, 08:58 PMThePerfectHacker
- January 29th 2007, 09:07 PMcalcidiots101
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, 10:48 PMqpmathelp
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, 11:22 PMticbol
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, 11:59 AMcalcidiots101
thanks so much for the help!