integration problemssss (1 Viewer)

[ToO LaZy ^*]

1.
i tried using the t-formula but couldn't get the answer out...am i using the right method, or should i use something else?

 

[ToO LaZy ^*]

2. how do you do this?

 

[Affinity]

easy
1.)
first do a sin^2(x) = (1/2)(1 - cos(2x)) subs.

then you let t = tan(x), dx = 1/(t^2 + 1) dt

the integral should equal

integral {t = 0 to sqrt(3)} 1/(9 - t^2) dt


2.)
sin2x = 2sin(x)cos(x)

then you let u = sin(x) ...

 

[ToO LaZy ^*]

ok..thx

 

[ToO LaZy ^*]

Originally posted by Affinity
easy
1.)

the integral should equal

integral {t = 0 to sqrt(3)} 1/(9 - t^2) dt

err..i can't get down to:
integral {t = 0 to sqrt(3)} 1/(9 - t^2) dt
...
.
.

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
err..i can't get down to:
integral {t = 0 to sqrt(3)} 1/(9 - t^2) dt

Notice that sin²x = (1 - cos 2x) / 2. Using this, you should get int (from 0 to pi / 3) 1 / (4 + 5cos2x) dx.
Now, let t = tan x, and show that dx = dt / (t² + 1)
When x = 0, t = 0, and when x = pi / 3, t = sqrt(3).
Thus, we now have int (from 0 to sqrt(3)) 1 / [4 + 5 * (1 - t²) / (1 + t²)] * dt / (t² + 1), by expressing cos2x in terms of t.
This tidies up to int (from 0 to sqrt(3)) 1 / (9 - t²) dt
which is, using partial fractions, (1 / 6) * [ln |(t + 3) / (3 - t)|] (from 0 to sqrt(3)),
which is (1 / 6) * ln(2 + sqrt(3))

 

[CM_Tutor]

For the second one, u = sin²x is a faster approach, as it then follows that du = sin2x dx

 

[ToO LaZy ^*]

ohhhh...i see...i see.
i think what got me confused was the cos2(x) bit..

then i saw that you were using t = tan(x) instead of t = tan(x/2) and that cleared everything up

thx champ

Originally posted by CM_Tutor
For the second one, u = sin²x is a faster approach, as it then follows that du = sin2x dx

ok i'll try that out

 

[ToO LaZy ^*]

wow...MUCH faster indeed

 

[ToO LaZy ^*]

can someone teach me how to do 'reduction' questions...
here's a q.

 

[CM_Tutor]

Part One: I_n = int (from 0 to 1) x(1 - x³)ⁿ dx, for n => 0, n an integer.
Show that I_n = [3n / (2 + 3n)] * I_n-1 for n => 1

I_n = int (from 0 to 1) x(1 - x³)ⁿ dx, for integers n => 0
= int (from 0 to 1) (1 - x³)ⁿ d(x² / 2)
= [(1 - x³)ⁿx² / 2] (from 0 to 1) - int (from 0 to 1) x² / 2 d(1 - x³)ⁿ
= [0 - 0] - int (from 0 to 1) (x² / 2) * n(1 - x³)^n-1 * -3x² dx, for integers n => 1
= (3n / 2) * int (from 0 to 1) x⁴(1 - x³)^n-1 dx
= -1 * (3n / 2) * int (from 0 to 1) x(1 - x³ - 1)(1 - x³)^n-1 dx
= -1 * (3n / 2) * int (from 0 to 1) x(1 - x³)(1 - x³)^n-1 - x(1 - x³)^n-1 dx
= -1 * (3n / 2) * int (from 0 to 1) x(1 - x³)ⁿ dx - -1 * (3n / 2) * int (from 0 to 1) x(1 - x³)^n-1 dx

So, we have I_n = -1 * (3n / 2) * I_n + (3n / 2) * I_n-1
(1 + 3n / 2)I_n = (3n / 2)I_n-1
(2 + 3n)I_n = 3n * I_n-1

So, I_n = [3n / (2 + 3n)] * I_n-1, for integers n => 1, as required.

Part Two: Find I_n in terms of n, for n=> 0

First, note that I₀ = int (from 0 to 1) x(1 - x³)⁰ dx = int (from 0 to 1) x dx
= [x² / 2] (from 0 to 1)
= (1 / 2) - 0
= 1 / 2

Now, I_n = [3n / (2 + 3n)] * I_n-1
= [3n / (3n + 2)] * [3(n - 1) / (2 + 3(n - 1))] * I_n-2
= [3n / (3n + 2)] * [3(n - 1) / (3n - 1)] * [3(n - 2) / (2 + 3(n - 2))]I_n-3
= [3n / (3n + 2)] * [3(n - 1) / (3n - 1)] * [3(n - 2) / (3n - 4)] * [3(n - 3) / (2 + 3(n - 3))]I_n-4
= [3n / (3n + 2)] * [3(n - 1) / (3n - 1)] * [3(n - 2) / (3n - 4)] * [3(n - 3) / (3n - 7)] * ... * I₁
= [3n / (3n + 2)] * [3(n - 1) / (3n - 1)] * [3(n - 2) / (3n - 4)] * [3(n - 3) / (3n - 7)] * ... * (3 / 5) * I₀
= [3n / (3n + 2)] * [3(n - 1) / (3n - 1)] * [3(n - 2) / (3n - 4)] * [3(n - 3) / (3n - 7)] * ... * (3 / 5) * (1 / 2)
= [3n * 3(n - 1) * 3(n - 2) * ... * 3] / [(3n + 2) * (3n - 1) * (3n - 4) * ... * 5 * 2]
= [3ⁿ * n * (n - 1) * (n - 2) * ... * 1] / [(3n + 2) * (3n - 1) * (3n - 4) * ... * 5 * 2]
= 3ⁿn! / [(3n + 2) * (3n - 1) * (3n - 4) * ... * 5 * 2]

Can anyone see a nice way to simplify the expression 2 * 5 * 8 * ... * (3n + 2)?

 

[ToO LaZy ^*]

whoooaa...holy shhiiivers..i hope this is easier than it looks, coz it gonna take me ages to disgest all that
but thatnks heaps for the help CM_tutor

 

[ToO LaZy ^*]

Originally posted by CM_Tutor

= 3ⁿn! / [(3n + 2) * (3n - 1) * (3n - 4) * ... * 5 * 2]

is that the final answer ? *crosses fingers*

 

[CM_Tutor]

Originally posted by ToO LaZy ^*
is that the final answer ? *crosses fingers*

Yes, unless someone suggests a nice way to simplify 2 * 5 * 8 * ... * (3n + 2).

 

[ToO LaZy ^*]

could someone give me a hint to how i would start these questions..

 

[George W. Bush]

1) Times both sides by sqrt 5-x, then complete the square on the bottom.

2) Times both sides by cosx, then note what the derivative of sinx + cosx is.

 

[KeypadSDM]

With the initial second question, isn't it just a straight integration to Ln[2 + sin^2[x]] + c? What's with all the substitution? Or am I hideously wrong?

 

[ToO LaZy ^*]

ok.
for 1, i broke down the integral into 2 parts:

integral{x=2 ->4} 5 / sqrt[(3-x)^2 - 14] dx - integral{x=2 ->4} x / sqrt[(3-x)^2 - 14] dx

how do you do this bit...integral{x=2 ->4} x / sqrt[(3-x)^2 - 14] dx ...?
or did i do something wrong to start with?...

 

[George W. Bush]

what is the derivative of (3-x)^2 - 14?

Manipulate it so that you get that on top..

 

[ToO LaZy ^*]

d/dx(3-x)^2 -14
= 2(3-x) * -1
= -2(3-x)
= -6+2x.....
and then what?...this is where i have most of my problems .

EDIT: or is it 6-2x..?

EDIToops...made a stupid mistake..(3-x)^2 -14 isn't the completed square -_-"..
let me try again