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?
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)
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.....