sucky integration question Watch

This discussion is closed.
It'sPhil...
Badges: 2
Rep:
?
#1
Report Thread starter 15 years ago
#1
Sorry talking about skool stuff in the holidays but my maths teacher was *kind* enough to set us 130 integration q's to do. Anywho i thought id get them out the way early and i got sorta stuck on this (Q77 Edexcel P5 p48)

If anyone could help id appreciate it (btw below S is integration symbol thing)

I = S _____dx________ (between 0 and 1)
.........x + sqrt(1 - x^2)

now i let x = sint => dx/dt = cost

=> I = S ___cost____ dt (between pi/2 and 0)
............. sint + cost

now i can't do this integral from here - the question ses using the substitution x = sint show that I = pi/4.

I cant integrate the above expression as is and need to make a further substitution u = tant then muck about with partial fraction and logs and eventually get a slightly fudged answer. Is the above easily intergatable??

please help thnx.
0
Unregistered
Badges:
#2
Report 15 years ago
#2
mate, i tried doing it but it's a while since a last done integration. What I think you have to do is try and simplify it as much as you can (which I think you have already done) then use integration by parts. Then you come out with an expression in which you'll probably require some identities and a logical deduction showing that I=pi/2.

Don't use logs - try keeping things as simple as possible e.g. don't create disorder out of order.

Let us know how you get on...
Rich
Badges: 2
Rep:
?
#3
Report 15 years ago
#3
(Original post by It'sPhil...)
Sorry talking about skool stuff in the holidays but my maths teacher was *kind* enough to set us 130 integration q's to do. Anywho i thought id get them out the way early and i got sorta stuck on this (Q77 Edexcel P5 p48)

If anyone could help id appreciate it (btw below S is integration symbol thing)

I = S _____dx________ (between 0 and 1)
.........x + sqrt(1 - x^2)

now i let x = sint => dx/dt = cost

=> I = S ___cost____ dt (between pi/2 and 0)
............. sint + cost

now i can't do this integral from here - the question ses using the substitution x = sint show that I = pi/4.

I cant integrate the above expression as is and need to make a further substitution u = tant then muck about with partial fraction and logs and eventually get a slightly fudged answer. Is the above easily intergatable??

please help thnx.
Hmm... tricky!

You need to notice that d/dt(ln(sin(t) + cos(t))) = (cos(t) - sin(t))/(sin(t) + cos(t)). You can also see that d/dt(t) = 1 = (sin(t) + cos(t))/(sin(t) + cos(t)), and if you add those together you get:

d/dt(ln(sin(t) + cos(t))) + d/dt(t) = (cos(t) - sin(t))/(sin(t) + cos(t)) + (sin(t) + cos(t))/(sin(t) + cos(t)) = (2cos(t))/(sin(t) + cos(t)), so then we just need to halve that, so:

I = 1/2(t + ln(sin(t) + cos(t))) + C

Now we substitute our t = arcsin(x) and we get:

I = 1/2(arcsin(x) + ln(x + cos(arcsin(x)))) + C

I'll leave you to put in the limits.

Hope this helps,
0
It'sPhil...
Badges: 2
Rep:
?
#4
Report Thread starter 15 years ago
#4
Cheers!! Very clever method!

THats much quicker than my original method wich involved arctans and partial fractions and the upper limit being infinity which seemed dodgy.

anyhow thanks for the help
0
Rich
Badges: 2
Rep:
?
#5
Report 15 years ago
#5
(Original post by It'sPhil...)
Cheers!! Very clever method!

THats much quicker than my original method wich involved arctans and partial fractions and the upper limit being infinity which seemed dodgy.

anyhow thanks for the help
Sounds fun, you should try it and see if you get a valid answer!
0
Jonny W
Badges:
#6
Report 15 years ago
#6
Here's a different approach, without the logs. Let

I = (int from 0 to pi/2) cos x / (sin x + cos x) dx
J = (int from 0 to pi/2) sin x / (sin x + cos x) dx

Substitute y = pi/2 - x into I and use

cos (pi/2 - y) = sin y
sin (pi/2 - y) = cos y

to get I = J.

But it's pretty obvious that I + J = pi/2.

So I = pi/4.

Jonny.
It'sPhil...
Badges: 2
Rep:
?
#7
Report Thread starter 15 years ago
#7
(Original post by rahaydenuk)
Sounds fun, you should try it and see if you get a valid answer!
the bizarre thing is that it does give pi/4!!

basically i write I as S[pi/2,0] 1/(tant + 1) dt

then let u = tant => dt/du = 1/(1 + u^2)

=> I = S[inf,0] 1/[(u+1)(1+u^2)] du

now use partial fractions and you end up with (i think)

I = (1/2)S[inf,0] 1/(u+1) + 1/(1+u^2) - u/(1+u^2) du

which hopefully comes out to be

[ 1/4 ln[(u^2+2u+1)/(1+u^2)] + 1/2 arctanu ] {inf,0}

both logs come out as ln 1 ie 0 and you end up with 1/2(pi/2) = pi/4

is this method vaguely correct? Apart from it being long and the improper integral it all seems valid (i hope!)
0
Surfing Hamster
Badges: 1
Rep:
?
#8
Report 15 years ago
#8
Woooo!!! Go Rich with the arc-tan-thingys!!!!
0
Rich
Badges: 2
Rep:
?
#9
Report 15 years ago
#9
(Original post by Jonny W)
Here's a different approach, without the logs. Let

I = (int from 0 to pi/2) cos x / (sin x + cos x) dx
J = (int from 0 to pi/2) sin x / (sin x + cos x) dx

Substitute y = pi/2 - x into I and use

cos (pi/2 - y) = sin y
sin (pi/2 - y) = cos y

to get I = J.

But it's pretty obvious that I + J = pi/2.

So I = pi/4.

Jonny.
One word, beautiful.

Thankyou!
0
Nylex
Badges: 10
Rep:
?
#10
Report 15 years ago
#10
Bah, if only it was P3.
0
Rich
Badges: 2
Rep:
?
#11
Report 15 years ago
#11
(Original post by It'sPhil...)
the bizarre thing is that it does give pi/4!!

basically i write I as S[pi/2,0] 1/(tant + 1) dt

then let u = tant => dt/du = 1/(1 + u^2)

=> I = S[inf,0] 1/[(u+1)(1+u^2)] du

now use partial fractions and you end up with (i think)

I = (1/2)S[inf,0] 1/(u+1) + 1/(1+u^2) - u/(1+u^2) du

which hopefully comes out to be

[ 1/4 ln[(u^2+2u+1)/(1+u^2)] + 1/2 arctanu ] {inf,0}

both logs come out as ln 1 ie 0 and you end up with 1/2(pi/2) = pi/4

is this method vaguely correct? Apart from it being long and the improper integral it all seems valid (i hope!)
Yes it appears to work!

I was just a bit worried about the infinities, as rigorously speaking, you can't just deal with them using finite algebra ... I'm not sure what would happen in exam conditions though.
0
Jonny W
Badges:
#12
Report 15 years ago
#12
If you're worried about infinities, you can integrate over a slightly smaller range and then take a limit.

Let

I = (int from 0 to pi/2) 1/(tan t + 1) dt

and for each x with 0 < x < pi/2 define

I(x) = (int from 0 to (pi/2 - x)) 1/(tan t + 1) dt.

You can work out I(x) using your u = tan t substituion (no infinities needed), then observe that

I(x) -> pi/4 as x -> 0.

Hence I = pi/4.

[According to my computer algebra program -- I'm too lazy to do it on paper --

I(x) = (pi - 2x + 2 log(cos(x) + sin(x)))/4.]

Jonny.
Unregistered
Badges:
#13
Report 15 years ago
#13
nice one Jonny and a very elegant solution (much more so than my long cumbersome one!!). Have you done the STEP question which is very similar (possibly in Siklos's book) or did you think of it yourself in a moment of inspiration??
Unregistered
Badges:
#14
Report 15 years ago
#14
(Original post by Unregistered)
nice one Jonny and a very elegant solution (much more so than my long cumbersome one!!). Have you done the STEP question which is very similar (possibly in Siklos's book) or did you think of it yourself in a moment of inspiration??
I probably got the idea from question 20 in Siklos's second book (http://www.maths.cam.ac.uk/undergrad/advancedproblems/) although that question is a bit harder.

Jonny.
Unregistered
Badges:
#15
Report 15 years ago
#15
[QUOTEI = S _____dx________ (between 0 and 1)
.........x + sqrt(1 - x^2)

now i let x = sint => dx/dt = cost

=> I = S ___cost____ dt (between pi/2 and 0)
............. sint + cost [/QUOTE]

The answer is: squirt
X
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Brexit: Given the chance now, would you vote leave or remain?

Remain (1411)
79.63%
Leave (361)
20.37%

Watched Threads

View All