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- Thread starter richz
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This one is a bit of a biatch.
J<sub>n</sub> = ∫ cos<sup>n-1</sup>xsin(nx) dx
let u = cos<sup>n-1</sup>x , u' = -(n-1)sinx.cos<sup>n-2</sup>x
v' = sin(nx) , v=(-1/n)cosnx
J<sub>n</sub> = (-1/n)cosnx.cos<sup>n-1</sup>x - (n-1)/n ∫ cos<sup>n-2</sup>xcos(nx)sinx
then, working with the second part
∫ cos<sup>n-2</sup>xcos(nx)sinx
= ∫ cos<sup>n-2</sup>xsinx[cos(n-1)xcosx - sin(n-1)xsinx]
= ∫ cos<sup>n-2</sup>xcos(n-1)xsinx - cos<sup>n-2</sup>xsin(n-1)xsin<sup>2</sup>x
= ∫ cos<sup>n-1</sup>x[sinxcos(n-1) + cosxsin(n-1)x] - cos<sup>n-2</sup>xsin(n-1)x
= ∫ cos<sup>n-1</sup>xsin(nx) - cos<sup>n-2</sup>xsin(n-1)x
= J<sub>n</sub> - J<sub>n-1</sub>
J<sub>n</sub> = (-1/n)cosnx.cos<sup>n-1</sup>x - (n-1)/n(J<sub>n</sub> - J<sub>n-1</sub>)
J<sub>n</sub>([2n-1]/n) = (1/n)[(n-1)J<sub>n-1</sub> - cosnx.cos<sup>n-1</sup>x]
J<sub>n</sub> = (1/2n-1)[(n-1)J<sub>n-1</sub> - cosnx.cos<sup>n-1</sup>x]
I swear I hit quote, not edit.
J<sub>n</sub> = ∫ cos<sup>n-1</sup>xsin(nx) dx
let u = cos<sup>n-1</sup>x , u' = -(n-1)sinx.cos<sup>n-2</sup>x
v' = sin(nx) , v=(-1/n)cosnx
J<sub>n</sub> = (-1/n)cosnx.cos<sup>n-1</sup>x - (n-1)/n ∫ cos<sup>n-2</sup>xcos(nx)sinx
then, working with the second part
∫ cos<sup>n-2</sup>xcos(nx)sinx
= ∫ cos<sup>n-2</sup>xsinx[cos(n-1)xcosx - sin(n-1)xsinx]
= ∫ cos<sup>n-2</sup>xcos(n-1)xsinx - cos<sup>n-2</sup>xsin(n-1)xsin<sup>2</sup>x
= ∫ cos<sup>n-1</sup>x[sinxcos(n-1) + cosxsin(n-1)x] - cos<sup>n-2</sup>xsin(n-1)x
= ∫ cos<sup>n-1</sup>xsin(nx) - cos<sup>n-2</sup>xsin(n-1)x
= J<sub>n</sub> - J<sub>n-1</sub>
J<sub>n</sub> = (-1/n)cosnx.cos<sup>n-1</sup>x - (n-1)/n(J<sub>n</sub> - J<sub>n-1</sub>)
J<sub>n</sub>([2n-1]/n) = (1/n)[(n-1)J<sub>n-1</sub> - cosnx.cos<sup>n-1</sup>x]
J<sub>n</sub> = (1/2n-1)[(n-1)J<sub>n-1</sub> - cosnx.cos<sup>n-1</sup>x]
I swear I hit quote, not edit.
Last edited:
FinalFantasy
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SNIPER RIFLE!!antwan2bu said:
they look very nasty, so yours free to claim all ye like
if you let A, B and C represent z<sub>1</sub>, z<sub>2</sub> and (z<sub>1</sub> + z<sub>2</sub>) respectively then:
OACB is a parallelogram since point C is made by producing vector OA at B.
(z<sub>1</sub> + z<sub>2</sub>) = 2i(z<sub>1</sub> - z<sub>2</sub>) hence the diagonal AB is perpendicular to OC (since multiplying by 'i' causes a 90 degree rotation)
∴ OACB is a rhombus.
Hence OA = OB and |z<sub>1</sub>| = |z<sub>2</sub>|
OACB is a parallelogram since point C is made by producing vector OA at B.
(z<sub>1</sub> + z<sub>2</sub>) = 2i(z<sub>1</sub> - z<sub>2</sub>) hence the diagonal AB is perpendicular to OC (since multiplying by 'i' causes a 90 degree rotation)
∴ OACB is a rhombus.
Hence OA = OB and |z<sub>1</sub>| = |z<sub>2</sub>|
I'm going to BS my geometrical properties for a second here but for iii) :
The diagonals of the rhombus are in a proportion of 2:1 and the bisect each other creating triangles with sides of a ratio 2:1.
∠COB = ∠COA (angles of rhombus proveable via similar triangles SAS)
Hence if ∠AOB = α then ∠AOC = α/2
Since the opposite and the adjacent sides to ∠AOC in the Right Δ OAX (where X is the point of intersection of the diagonals) are in the proportion 2:1 then:
tanα/2 = 1/2
(you can probably do it with a fair bit less explanation)
...This BOS lag is killing me.
The diagonals of the rhombus are in a proportion of 2:1 and the bisect each other creating triangles with sides of a ratio 2:1.
∠COB = ∠COA (angles of rhombus proveable via similar triangles SAS)
Hence if ∠AOB = α then ∠AOC = α/2
Since the opposite and the adjacent sides to ∠AOC in the Right Δ OAX (where X is the point of intersection of the diagonals) are in the proportion 2:1 then:
tanα/2 = 1/2
(you can probably do it with a fair bit less explanation)
...This BOS lag is killing me.
Last edited:
iv)
tan(α ) = 2tan(α/2)/(1 - tan<sup>2</sup>[α/2]) = 1/(1-1/4)
tan(α ) = 4/3 ---> α = tan<sup>-1</sup>(4/3)
if z<sub>2</sub> = Kz<sub>1</sub> where K = x + iy
y/x = 4/3 ---> y = (4/3)x
|K|=1 ---> x<sup>2</sup> + y<sup>2</sup> = 1
x<sup>2</sup>(1 + 16/9) = 1
x<sup>2</sup> = 9/25 --> x = 3/5 so y = 4/5
K = 1/5(3 + 4i)
tan(α ) = 2tan(α/2)/(1 - tan<sup>2</sup>[α/2]) = 1/(1-1/4)
tan(α ) = 4/3 ---> α = tan<sup>-1</sup>(4/3)
if z<sub>2</sub> = Kz<sub>1</sub> where K = x + iy
y/x = 4/3 ---> y = (4/3)x
|K|=1 ---> x<sup>2</sup> + y<sup>2</sup> = 1
x<sup>2</sup>(1 + 16/9) = 1
x<sup>2</sup> = 9/25 --> x = 3/5 so y = 4/5
K = 1/5(3 + 4i)
Dumsum
has a large Member;
If you read the question you'll find that you don't have to prove the reduction formula...
Sorry about the delay, I'd tried to post a message the other night but I geuss it didn't make it through the BOS lag so I'll try and remember what I wrote.xrtzx said:
For the above question I could be pretty sure that int. by parts is qhat was required and the two parts I chose were the ones which seemed most obvious. After choosing how to split it up I was quickly able to see that it was going to produce a 'cosnx.cos<sup>n-1</sup>x' term which is part of the answer so I figured it was the right way to go. The integral I was working with quickly yielded: cos<sup>n-2</sup>xsin(n-1)xsin<sup>2</sup>x and after a quick inspection you will realise that turning the sin<sup>2</sup>x into (1 - cos<sup>2</sup>x) will give you a J<sub>n-1</sub> term. Once I had isolated that I figured that I just had to manipulate the rest of what was left to get a J<sub>n</sub> term to move over.
In terms of general tips, here is the best I can come up with at the moment:
- More than anything read the damn question or you'll end up doing what I did and going through a whole lot of unnecesary working when they're handing it to you on a silver platter.
- Most reduction questions tell you what you're trying to show. This is an invaluable reference point. If you look at the way I did the question above, it basically all hinges on manipulating parts of the expression to get it closer to the solution.
- watch out for simple tricks where you might not have to use integration by parts especially when all you need is some simple trig manipulation eg. ∫tan<sup>n</sup>x dx which can be made incredibly easy using trig manipulation.
- A common trick which comes up is where you have something of the form ∫ (x<sup>2</sup>+1)<sup>n-1</sup>.P(x) dx where you want to make up another power to get a power of 'n' for an I<sup>n</sup> term. In this case you can simply add and subtract x<sup>2</sup>(x<sup>2</sup>+1)<sup>n-1</sup>.P(x) so you get an I<sup>n</sup> term and something else which you have to play around with.
I hope that was the kind of stuff you were looking for.