SydneyRoverP6B said:
Doing some first year University Engineering Mathematics revision, somewhat stuck at the moment on the following two. Any pointers will be most welcome
The first is indefinite, the integrand is 3dx /( x ( lnx - (lnx)^2 ) ^1/2 )
The answer is 3 arcsin ( lnx - 1/2 ) + C
The second is definite with limits from 1 to 3 root 3. The integrand is dy / ( y^2/3 + y^4/3 )
The answer is PI / 4.
Ron.
I'm also scratching my head over the 2nd one.
I've had an answer which ALMOST agrees with the first one, a slight discrepancy towards the end though... :x
There are two things to remember here (well, three in a way...!)
I've had a "work backwards" moment, and thought about making the substitution that q = [ln (x) - 1/2], so dq = (1/x) dx, which takes care of the x in the denominator of your integrand!
q^2 = [ln (x) - 1/2]^2 = [{ln(x)}^2 - ln (x) + 1/4], seeing your integrand has ln(x) - {ln(x)}^2 in the denominator, is equal to (1/4) - q^2
3 dx / x [(ln(x) - {ln(x)}^2)^1/2] = 3 dq/[({1/4} - q^2)^1/2] = 6 dq/[(1 - 4q^2)^1/2] = 6 dq/[(1 - {2q}^2)^1/2]
(since we've multiplied by 4 within the square root, that equates to multiplying by 2 (the main square root of 4) when not within the square root i.e. s/a^(1/2) = 2s/{4a}^(1/2))
y = arcsin (kx) ==> sin (y) = kx ==> cos (y) dy/dx = k ==> dy/dx = k/[cos (y)], but remember since [sin(y)]^2 + [cos(y)]^2 = 1,
cos (y) = {1 - [sin(y)]^2}^(1/2) = {1 - (kx)^2}^(1/2)
Therefore if y = arcsin (kx) ==> dy/dx = k/[{1 - (kx)^2}^(1/2)]. (Get out clause...) It can also be shown that if y = arcos (kx) ==> dy/dx = -k/[{1 - (kx)^2}^(1/2)]
6 dq/[(1 - {2q}^2)^1/2], comparing this, we get that k = 2,
6 dq/[(1 - {2q}^2)^1/2] = 3 X (2 dq/[(1 - {2q}^2)^1/2])
Therefore 3 arcsin (2q) + C, since q = [ln (x) - 1/2], 2q = [2ln (x) - 1] = [ln (x^2) - 1]
So I get the integral to be
3 arcsin (2ln (x) - 1) + C or
3 arcsin (ln (x^2) - 1) + C
or, to be complete/awkward...
D - 3 arcos (2ln (x) - 1) or
D - 3 arcos (ln (x^2) - 1)
I think that's right because arcos (x) + arcsin (x) = PI/2 (i.e. a constant)