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# P5 questions help, thread:) watch

1. how do you do this?

integrate (7x-x^2)/(2-x)(x^2+1) dx between 0 and 1?

pk

2. they want me to show its equal to (7/2)ln2 - pi/4, but i dont know how to integrate thi other than partial fractions - and its taking forever, andyone know a shortcut?
3. rewrite the numerator as

-(1+x²) -7(2-x) + 15

and you end up with frractions and also do partial fractions.

its not tooo long!
4. (Original post by Fermat)
rewrite the numerator as

-(1+x²) -7(2-x) + 15

and you end up with frractions and also do partial fractions.

its not tooo long!
how'd you do that numerator thing - never been taught that; also, where do you go from there - the concurrent x^2 term in teh denominator is annoying me) - how do i deal with it?

also, how do you integrate 15/(2-x)(x^2 +1) that you get at the end?

must be a short cut

thanks, pk
5. (7x-x²) / (2-x)(x²+1) = A/(2-x) + (Bx+C)/(x²+1)
(7x-x²) = A(x²+1) + (Bx+C)(2-x)
x = 2: 10 = 5A => A=2
Compare constants: 0 = A + 2C => C=-1
Compare x² coefficient: -1 = A - B => B=3

(7x-x²) / (2-x)(x²+1) = 2/(2-x) + (3x-1)/(x²+1)

∫ 2/(2-x) + (3x-1)/(x²+1) dx
= ∫ 2/(2-x)dx + ∫3x/(x²+1)dx - ∫1/(x²+1)dx
= -2ln(2-x) + (3/2)ln(x²+1) - arctan(x) + c

(edit... silly me )
6. (Original post by SsEe)
(7x-x²) / (2-x)(x²+1) = A/(2-x) + (Bx+C)/(x²+1)
(7x-x²) = A(x²+1) + (Bx+C)(2-x)
...
that's simpler!

(Original post by SsEe)
...
So ∫2/(2-x) + (3x-1)/(x²+1) dx
= -2ln(2-x) -arctan(x) - (3/2)ln(x²+1) + c
That's what I got
7. (Original post by SsEe)
(7x-x²) / (2-x)(x²+1) = A/(2-x) + (Bx+C)/(x²+1)
(7x-x²) = A(x²+1) + (Bx+C)(2-x)
x = 2: 10 = 5A => A=2
Compare constants: 0 = A + 2C => C=-1
Compare x² coefficient: -1 = A - B => B=3

(7x-x²) / (2-x)(x²+1) = 2/(2-x) + (3x-1)/(x²+1)

∫ 2/(2-x) + (3x-1)/(x²+1) dx
= ∫ 2/(2-x)dx + ∫3x/(x²+1)dx - ∫1/(x²+1)dx
= -2ln(2-x) + (3/2)ln(x²+1) - arctan(x) + c

(edit... silly me )
did that but got my constants wrong, lol - no wonder i was going wrong
8. stuck on another few questions!!!

integral of sec^n dx between 0 and pi/3.. The resuction formula i got was

In = (2n-2) .(sqrt 3)/n-1 - (n-1)/(n-2) . In-2

Had to work out I7 - i keep on getting 103sqrt3 / 24 - (5/16) ln(2+sqrt 3), but the book has 61/8 rt 4 frp sp,e reason.

help

also stuck on these:
integrate 1/(x+sqrt(1-x2) dx, using x=sint
finally integrate sinx.(1+cos x) / (1-cosx) dx between pi/3 and pi/2
thanks

pk
9. For the last one, maybe try a t = tan(x/2) substitution...

I've kinda rushed it. But if I've done it right, then you get: (4) INT 1/[t(t2+1)] dt, which you could do partials on, and see how it goes.
10. (Original post by mockel)
For the last one, maybe try a t = tan(x/2) substitution...

I've kinda rushed it. But if I've done it right, then you get: (4) INT 1/[t(t2+1)] dt, which you could do partials on, and see how it goes.
1/[t(t2+1)] - kept on getting ->1/[t(t2+1)2]

which i couldmn't integrate - i'll check my working again

another question i'm having probs with is to show

integral sin n x / sin x

In= 2sin(n-1)x / (n-1) +In-2

dont know what step to start with - -tried compound angle formula and taking the sin up and parts, but not getting anywhere - hope they give us and easy integrate (xn sin x)dx one in the exam
11. Yeah, sorry, it definitely is what you said. I'll think about it some more.
12. (Original post by mockel)
Yeah, sorry, it definitely is what you said. I'll think about it some more.
what a weird question huh- out of all the papers i've dont i've never come across such awkward integrals - these are from the exercises in the book - never touched um till yesterday

pk
13. The sinnx one, I remember doing. Didn't they say something like 'by considering I2 - In'?
14. (Original post by mockel)
The sinnx one, I remember doing. Didn't they say something like 'by considering I2 - In'?
if you gopt P5 edexcell - its from pg 55 at the top - doesn't give any hints - just say to show its that reduction thing,

pk
15. For the sin(nx)/sinx one, note that:
sin(nx) - sin[(n-2)x] = 2cos[(n-1)x]sinx

Therefore:
I(n) = ∫ (2cos[(n-1)x]sinx + sin[(n-2)x])/sinx) dx
= ∫ 2cos[(n-1)x] dx + ∫ sin[(n-2)x]/sinx dx
= [2sin(n-1)x]/(n-1) + I(n-2)

integrate 1/(x+sqrt(1-x2) dx, using x=sint
This is horrible! Are you sure you typed it out right?
16. it was integral 1/(x+sqrt(1-x2)) using x=sin t, lol

dont know how its supposed to help; i got integral of cos t/(sint + cos t), and didn't know where to progress from there

pk
17. (Original post by dvs)
For the sin(nx)/sinx one, note that:
sin(nx) - sin[(n-2)x] = 2cos[(n-1)x]sinx
could you explain that line???

how the hell are we suppoed to know that, lol - would never have thought of that in an exam - did you know it from practice, or are you a wizz with trig identities?

figured out you used sum-product, but HOW did you spot that - starting to get worried about tomorrow - may be harder than i first though

pk
18. Crazy. Anyway, I remember the hint from the paper now, and it was by considering In-2 - In:

In-2 - In = ∫ [sin(n+2)x - sin(n)x / sinx] dx

Using the identity, sinA - sinB = 2cos[(A+B)/2]sin[(A-B)/2]

In-2 - In = ∫ 2[cos(n+1)x.sinx] / sinx dx

In-2 - In = (2) ∫ cos(n+1)x dx

In-2 - In = 2sin(n+1)x/(n+1)

=> In = In-2 - 2sin(n+1)x/(n+1)

Argh, too late. Should've done it the other way as well. If I delete it, it'll seem like a waste of time. So, I won't. =D
19. (Original post by mockel)
Crazy. Anyway, I remember the hint from the paper now, and it was by considering In-2 - In:

In-2 - In = ∫ [sin(n+2)x - sin(n)x / sinx] dx

Using the identity, sinA - sinB = 2cos[(A+B)/2]sin[(A-B)/2]

In-2 - In = ∫ 2[cos(n+1)x.sinx] / sinx dx

In-2 - In = (2) ∫ cos(n+1)x dx

In-2 - In = 2sin(n+1)x/(n+1)

=> In = In-2 - 2sin(n+1)x/(n+1)

Well isn't that strange. Probably made a mistake somewhere along the line...

Argh, too late. Should've done it the other way as well. If I delete it, it'll seem like a waste of time. So, I won't. =D
its n-2 - did they give that in the paper? - in the textbook, there are no hints, and i've no idea how we are supposed to 'see that'
20. Definitely gave the hint in the paper.

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