Two Integrals

SydneyRoverP6B

Well-Known Member
Staff member
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.
 
well youve reminded me that i used to know this stuff ... can i remember it...? erm....
 
On no, dead hard sums!. Sorry Ron, bit like Rich, it's been too long for me. The last uni maths exam I sat was back in the last century :shock: .
Maybe you could try an online Maths help forum? Found this one which seems like it might be worth a try http://mathhelpboards.com/ :?:
 
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)
 
darth sidious said:
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

Ron.

On checking, I cannot get 3 arcsin {ln(x) - 1/2} + C to differentiate into 3/( x [ln(x) - {ln(x)}^2 ) ^1/2 )

Are you sure that's the answer it's meant to be?
 
Thanks Steve for the link :) Just thought I'd throw it on here first, as I know Darth is more than good with maths.

Appreciate your reply Ant :)

Hmmm, my initial approach was to try and write the integrand in a recognised standard form, which for f(x) = arcsin(x) is dx / ( ( 1 - x^2)^1/2 ). Factoring an ln(x) out would go part of the way, but not the whole way.

Did you differentiate (implicitly) your answer Ant to see what the difference was in the integrands?

Ron.

Update,...yes, but I'll double check the book now.
 
Hi Ant,

Yes 3 arcsin ( ln(x) - 1/2 ) + C is the given answer. Doesn't mean it is definitely right though.

Ron.
 
darth sidious said:
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)

:shock: :shock:
 
Hi Ant,

I differentiated the text answer, and I don't achieve the integrand in the question either. :? Substituting in a value for x gives two totally different answers. Replacing 1/2 with 1 makes almost no difference at all in terms of changing the value obtained.

Ron.
 
SydneyRoverP6B said:
Thanks Steve for the link :) Just thought I'd throw it on here first, as I know Darth is more than good with maths.

Appreciate your reply Ant :)

Hmmm, my initial approach was to try and write the integrand in a recognised standard form, which for f(x) = arcsin(x) is dx / ( ( 1 - x^2)^1/2 ). Factoring an ln(x) out would go part of the way, but not the whole way.

Did you differentiate (implicitly) your answer Ant to see what the difference was in the integrands?

Ron.

Update,...yes, but I'll double check the book now.

No, my answer seems to work out...

y = 3 arcsin (2ln (x) - 1) + C, therefore (y - C) / 3 = arcsin (2ln (x) - 1), therefore sin [(y - C) / 3] = 2 ln (x) - 1

{(1/3) X cos [(y - C) / 3]} dy/dx = 2/x, or dy/dx = 6/(x {cos [(y - C) / 3]})

cos [(y - C) / 3] = (1 - {sin [(y - C) / 3]^2})^(1/2) = (1 - [2 ln (x) - 1]^2)^(1/2)

[2 ln (x) - 1]^2 = 4 (ln (x))^2 - 4 ln (x) + 1, so...

(1 - [2 ln (x) - 1]^2) = 1 - {4 (ln (x))^2 - 4 ln (x) + 1} = 1 - 4 (ln (x))^2 + 4 ln (x) - 1 = -4 (ln (x))^2 + 4 ln (x) = 4 [ln (x) - (ln (x))^2]

(1 - [2 ln (x) - 1]^2)^(1/2) = (4 [(ln (x) - (ln (x))^2])^(1/2)) = 2 [(ln (x) - (ln (x))^2]^(1/2)

dy/dx = 6/(x {cos [(y - C) / 3]}) = 6/(x {(1 - [2 ln (x) - 1]^2)^(1/2)}) = 6/(x {2 [(ln (x) - (ln (x))^2]^(1/2)}) = 3/(x [(ln (x) - (ln (x))^2]^(1/2))

dy/dx = 3/(x [(ln (x) - (ln (x))^2]^(1/2)), which I think is what you originally had!

Trying the one given in the book:-

y = 3 arcsin (ln (x) - 1/2) + C, therefore (y - C) / 3 = arcsin (ln (x) - 1/2), therefore sin [(y - C) / 3] = ln (x) - 1/2

{(1/3) X cos [(y - C) / 3]} dy/dx = 1/x, or dy/dx = 3/(x {cos [(y - C) / 3]})

cos [(y - C) / 3] = (1 - {sin [(y - C) / 3]^2})^(1/2) = (1 - [ln (x) - 1/2]^2)^(1/2)

[ln (x) - 1/2]^2 = (ln (x))^2 - ln (x) + 1/4, so...

(1 - [ln (x) - 1/2]^2) = 1 - {(ln (x))^2 - ln (x) + 1/4} = 1 - (ln (x))^2 + ln (x) - 1/4 = 3/4 - (ln (x))^2 + ln (x) = 3/4 + ln (x) - (ln (x))^2

(1 - [ln (x) - 1/2]^2)^(1/2) = (3/4 + ln (x) - (ln (x))^2)

dy/dx = 3/(x {cos [(y - C) / 3]}) = 3/(x {(1 - [ln (x) - 1/2]^2)^(1/2)}) = 3/(x [(3/4 + ln (x) - (ln (x))^2)]^(1/2)})

If y = 3 arcsin (ln (x) - 1/2) + C, then dy/dx = 3/(x [(3/4 + ln (x) - (ln (x))^2)]^(1/2)}), which is not the integrand given!

I'm 99.99999% certain the author of the book has made a substitution, but forgotten to double it at the end!
 
I think I have the second one too...

dy / [y^(2/3) + y^(4/3)] = dy / [y^(2/3) {1 + y^(2/3)}] = y^(-2/3) dy / [1 + y^(2/3)] (i.e. factor out y^(2/3) from the bottom, and bring it to the top as y^(-2/3) )

Put q = 3 y^(1/3), so dq = y^(-2/3) dy (which looks like our numerator!)

and therefore y^(2/3) = (q/3)^2

the integrand becomes dq / [1 + (q/3)^2]

Let's make ANOTHER substitution! put q/3 = sinh (λ), or q = 3 sinh (λ), dq = 3 cosh (λ) dλ

Since ({cosh (λ)}^2 - {sinh (λ)}^2) = 1, we can say [1 + (q/3)^2] = 1 + {sinh (λ)}^2 = {cosh (λ)}^2

The integrand becomes {[3 cosh (λ)] dλ / [cosh (λ)]^2} dλ = {3 dλ / cosh (λ)} = 3 sech (λ) dλ.

I'm told that the integral of sech (λ) is the gudermannian (named after Christoph Gudermann, 1798–1852), and is defined as gd(x) = arctan {sinh(x)}

Therefore, the integral of 3 sech (λ) dλ is 3 arctan {sinh(λ)} + C

But since q/3 = sinh (λ), we can say 3 arctan {q/3} + C

Since y^(2/3) = (q/3)^2, y^(1/3) = (q/3), we can also say 3 arctan {y^(1/3)} + C

Seeing this is a definite integral (i.e. we have limits of integration), we can dispense with the constant C

The integral becomes 3 arctan {y^(1/3)} between 1 and 3 root (3), = (3 arctan {y^(1/3)} @ y = 3 root (3)) - (3 arctan {y^(1/3)} @ y = 1) = 3 [(arctan {y^(1/3)} @ y = 3 root (3)) - (arctan {y^(1/3)} @ y = 1)]

3 root 3 = (root (3)) cubed = (root (3)) ^ 3 = 3^(3/2), so when y = 3^(3/2), then y^(1/3) = [3^(3/2)]^(1/3) = 3^(1/2) = root (3)

i.e. y^(1/3) = root (3)

When y = 1, y^(1/3) = 1 (principal result, ignoring the complex conjugate pair results!)

arctan (root (3)) = PI/3 (radians) (because sin (PI/3) = root (3) / 2, cos (PI/3) = 1/2, therefore tan (PI/3) = sin (PI/3) / cos (PI/3) = {[root (3) / 2] / [1/2]} = root (3))

arctan (1) = PI/4 (radians) (because sin (PI/4) = cos (PI/4) = 1/ root (2), therefore tan (PI/4) = sin (PI/4) / cos (PI/4) = {[1/ root (2)] / [1/ root (2)]} = 1)

The definite integral = 3 X (PI/3 - PI/4) = 3 X PI ([1/3] - [1/4]) = 3 X PI/12 = PI/4

([1/3] - [1/4] = [4/12] - [3/12] = [1/12], of course)

The result of the definite integral is PI/4

Which, for this time, I agree with the book you have Ron!! :LOL:
 
Awesome Ant :D

Yes. the text answer for the first integral is wrong. I differentiated (implicitly) your answer, y = 3 arcsin (2lnx - 1), and achieved the integrand in the question also.

I'll have a look at your methodology for the second integral. :)

Ron.
 
Hi Ant,

I have come up with another alternative solution to the first integral. Sorry but I am also wavering on the substitutions that you used for both. I agree with PI/4 for the second integral, although my method was quite different.

I'll send you a PM.

Ron.
 
DaveHerns wrote,...
I got as far as differentiation and integration in A level Maths and after that nothing made any sense

It is all to do with the application, which is rarely explored at school.

Ron.
 
SydneyRoverP6B said:
DaveHerns wrote,...
I got as far as differentiation and integration in A level Maths and after that nothing made any sense

It is all to do with the application, which is rarely explored at school.

Ron.

This thread has turned into the Ron and Ant show! :oops: :roll: :LOL: Sorry people! :mrgreen:
 
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