# sucky integration questionWatch

This discussion is closed.
#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??

0
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...
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??

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
#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
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
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.
#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
15 years ago
#8
Woooo!!! Go Rich with the arc-tan-thingys!!!!
0
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
15 years ago
#10
Bah, if only it was P3.
0
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
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.
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??
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.
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]

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