Would you like to do some challenging integration questions? (1 Viewer)

[.ben]

1.

dx
∫ ---------------
(4x³-3x)sqrt(1-x²)

2.

Let t_n=∫[pi/2-->0]xsinⁿxdx
Prove that t_n=1/n²+[(n-1)/n]*t_n-2

3.

Prove I_n=∫[pi/2-->0]cos²ⁿxdx=[pi(2n)!/2²ⁿ⁺¹(n!)²]

4.

Given that If I_2n+1=∫[1-->0]x²ⁿ⁺¹e^x2dx
Prove that
I_2n+1=e/2-nI_2n-1


That's all for now thanks.

 

[_ShiFTy_]

Hmm...

Q1...confusing me too much

Q2 - took me some time to get the first step right but finally got it!

Q3 - wtf... i see factorial stuff, i run away and hide

Q4 - similar to Q4 but easier to spot


thx for these questions ben, might come in handy for my assessment tmrw

 

[.ben]

Lol i have an assessment tomorrow as well!

 

[_ShiFTy_]

Hehe, good luck!

btw, do you have the answer to Q1...and any hints for Q3?

 

[.ben]

yeh you too

sorry don't have any of the answers.

 

[Yip]

Shifty:
ara~.~ didnt notice the reciprocal -_- rather dodgy typesetting
q3: Obtain a reduction formula for the integral by taking the cosx part out and itnegrate it, then apply it a billion times

 

[_ShiFTy_]

Shifty:
q1: factorize x out and make substitution t^2=x^2-1, then it should come out pretty quickly

But then you need a 2x.dx on top, which we dont have

 

[.ben]

I fed question 1 to integrator. it came out with a really long log expression??!?!

 

[hyparzero]

There are 5 log terms actually, and without an elegant method, it would take approximately 3 pages to work it out

∫(1/((4*x^3-3*x)*sqrt(1-x^2)), x) =

1/3*ln(3*tan(1/2*arcsin(x))-sqrt(3))-1/3*ln(-3*tan(1/2*arcsin(x))+3*sqrt(3))
-1/3*ln(3*tan(1/2*arcsin(x))+3*sqrt(3))+1/3*ln(-3*tan(1/2*arcsin(x))-sqrt(3))-1/3*ln(1/x-sqrt(1-x^2)/x) + C

 

[Riviet]

I wouldn't say this is challenging, but it certainly is an interesting one that I posted last week:
http://community.boredofstudies.org/newreply.php?do=newreply&noquote=1&p=2389149

P.S Read my latest post in the thread for the "the clever substitution" hint.

 

[_ShiFTy_]

Just a general question about Q3, how can you get factorials from doing integration involving reduction?

 

[who_loves_maths]

1.

_________________dx
∫ __________________________________
__(4x³-3x)sqrt(1-x²)

2.

Let t_n=∫[pi/2-->0]xsinⁿxdx
Prove that t_n=1/n²+[(n-1)/n]*t_n-2

3.

Prove I_n=∫[pi/2-->0]cos²ⁿxdx=[pi(2n)!/2²ⁿ⁺¹(n!)²]

4.

Given that If I_2n+1=∫[1-->0]x²ⁿ⁺¹e^x2dx
Prove that
I_2n+1=e/2-nI_2n-1

Hi .ben, hope this isn't too late of a reply if you actually need(ed) solutions to any of these questions.

QUESTION 1: Find ∫dx/[x(4x² -3).Sqrt(1-x²)]

Make the substitution Sqrt(1-x²) = u ; -xdx/Sqrt(1-x²) = du

I = -∫du/(1-u²)(1-4u²)

then use the method of partial fractions to obtain :

I = (1/6)∫du/(1+u) + (1/6)∫du/(1-u) - (2/3)∫du/(1+2u) - (2/3)∫du/(1-2u)
= (1/6)ln|1+u| - (1/6)ln|1-u| - (1/3)ln|1+2u| + (1/3)ln|1-2u| + C
= (1/6)ln|(1+u)/(1-u)| + (1/3)ln|(1-2u)/(1+2u)| + C

Alternatively, you can also use standard integrals to assist during, say, an exam.


QUESTION 2: Show I_n = 1/n² + ((n -1)/n)I_n-2

I_n = {pi/2 -> 0}∫xSinⁿx dx = ∫(xSin^n-2x)Sin²x dx

where Sin²(t) = 1 - Cos²(t) :

I_n = ∫xSin^n-2x dx - ∫xSin^n-2x Cos²x dx

---> ∫xSin^n-2x Cos²x dx = I_n-2 - I_n ____________________ (1)

Now, apply integration by parts to the original integral I_n:

let u = xSin^n-1x , and dv = Sin(x) dx

you get:

I_n = [-xSin^n-1x Cos(x)]{pi/2 -> 0} + ∫Cos(x)[(n-1)xSin^n-2x Cos(x) + Sin^n-1x] dx

but [-xSin^n-1x Cos(x)]{pi/2 -> 0} = 0

I_n = (n-1)∫xSin^n-2x Cos²x dx + ∫Sin^n-1x Cos(x) dx

but {pi/2 -> 0}∫Sin^n-1x Cos(x) dx = (1/n)[Sinⁿx]{pi/2 -> 0}= 1/n

i.e. I_n = (n-1)∫xSin^n-2x Cos²x dx + 1/n ____________________ (2)

Finally, substitute (1) into (2) :

I_n = (n-1)(I_n-2 - I_n) + 1/n

---> I_n = 1/n^2 + ((n -1)/n)I_n-2


QUESTION 3: Show I_n = (pi.(2n)!)/(2²ⁿ⁺¹ n!²)

Proceed by induction.

1) n = 1
LHS = I₁ = {pi/2 -> 0}∫Cos²x dx = pi/4 = RHS.
n = 1 true.

2) Assume truth for n = k; to show for n = (k+1);

3) Consider:

I_k+1 := i_2(k+1) and I_k := i_2k ; where ":=" denotes "is equivalent to", and i_m denotes the m-th term of the recurrence integral: i_m = ∫Cos^mx dx

We proceed to use the reduction formula for the Cosine recurrence integral:

i_m = [(1/m)Cos^m-1x Sin(x)]{pi/2 -> 0} + ((m-1)/m).i_m-2

but [(1/m)Cos^m-1x Sin(x)]{pi/2 -> 0} = 0

---> i_m = ((m-1)/m).i_m-2

So we have:

I_k+1 = i_2k+2 = ((2k+1)/(2k+2)).i_2k = ((2k+1)/(2k+2))I_k

Using our initial assumption for n = k:

I_k+1 = ((2k+1)/(2k+2))[(pi.(2k)!)/(2^2k+1 k!²)]

multiply top and bottom by (2k+2) to obtain:

I_k+1 = [(2k+1)(2k+2)/(2k+2)²].[(pi.(2k)!)/(2^2k+1 k!²)]
= (pi.(2(k+1))!)/(2^2(k+1)+1 (k+1)!²)

Therefore, the case of n = (k+1) is true when n = k is true.

This completes the proof.


QUESTION 4: Prove I_2n+1 = e/2 - 2nI_2n-1

I_2n+1 = {1 -> 0}∫x²ⁿ⁺¹ e^x2 dx = (1/2)∫(x²ⁿ(2xe^x2)) dx

Proceed via integration by parts:

let u = x²ⁿ , and dv = 2xe^x2 dx ---> du = 2nx^2n-1 , and v = e^x2

I_2n+1 = (1/2)(x²ⁿ e^x2){1 -> 0} - 2n∫x^2n-1 e^x2 dx

but (x²ⁿ e^x2){1 -> 0} = e , and ∫x^2n-1 e^x2 dx = I_2n-1

-----> I_2n+1 = e/2 - 2nI_2n-1


Hope this may help, cheers.


P.S.1. I believe there is a 2 in front of the 'nI_2n-1' term in Question 4.

P.S.2. Apologies for any incorrect/unsightly typesetting.

 

[Slidey]

This forum really neds TeX.

 

[who_loves_maths]

You don't say! haha.