HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

Davo_01 · Apr 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
Let's generalize that just a little bit

$\\ $Find the re-currence formula in$ \ n \ $for$ \\ \\ I_n = \int_0^1 x^n (1-x^m)^{1/k} \ dx \ \ $for$ \ n > (m+1) \ $for naturals$ \ n$

(its doable I tried it)

Very nice question

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Let u= x^(n-2) du = (n-2)x^(n-3)
dv = x^2 (sqrt(1-x^3)) v= -(2/9) (1-x^3)^(3/2)

Un= (-(2/9)(1-x^3)^(3/2)x^(n-2))_1_0 +(2/9)(n-2) integral_1_0 ((1-x^3)^(3/2)(x^(n-3))
= 0 + 2/9 (n-2) (2/9)(n-2) (integral_1_0 ( (1-x^3)^(1/2))(1-x^3) x^(n-2)))
=(2/9)(n-2)[ integral _1_0 ((1-x^3)^(1/2) (x^(n-3)) - integral _1_0 (x^(n) ((1-x^3)^(1/2))]
=2/9 (n-2) (U_(n-3)-U_(n))
Rearranging (taking the U_n to the other side with its coefficient)
((2n+3)/9)U_(n)= (2/9)(n-2) (U_(n-3))
Simplifying U_n = ((2n-4)/(2n+3)) U_(n-3))
but n is not greater than or equal to 4. since U_(n-3) and n-3 > 0 but I dont know why it is greater than or equal to 5.
Sorry if you had to take a long time to read it (if you did read it).

Davo_01 · Apr 10, 2014

Re: MX2 Integration Marathon

hit patel said:
Let u= x^(n-2) du = (n-2)x^(n-3)
dv = x^2 (sqrt(1-x^3)) v= -(2/9) (1-x^3)^(3/2)

Un= (-(2/9)(1-x^3)^(3/2)x^(n-2))_1_0 +(2/9)(n-2) integral_1_0 ((1-x^3)^(3/2)(x^(n-3))
= 0 + 2/9 (n-2) (2/9)(n-2) (integral_1_0 ( (1-x^3)^(1/2))(1-x^3) x^(n-2)))
=(2/9)(n-2)[ integral _1_0 ((1-x^3)^(1/2) (x^(n-3)) - integral _1_0 (x^(n) ((1-x^3)^(1/2))]
=2/9 (n-2) (U_(n-3)-U_(n))
Rearranging (taking the U_n to the other side with its coefficient)
((2n+3)/9)U_(n)= (2/9)(n-2) (U_(n-3))
Simplifying U_n = ((2n-4)/(2n+3)) U_(n-3))
but n is not greater than or equal to 4. since U_(n-3) and n-3 > 0 but I dont know why it is greater than or equal to 5.
Sorry if you had to take a long time to read it (if you did read it).

On the 8th line u made a tiny mistake, 1+2(n-2)/9=(2n+5)/9 instead of (2n+3)/9

As for the n greater than equal to 5, it is not as easy to integrate I_1 as compared to integrating I_2, thats probably why.

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Davo_01 said:
On the 8th line u made a tiny mistake, 1+2(n-2)/9=(2n+5)/9 instead of (2n+3)/9

As for the n greater than equal to 5, it is not as easy to integrate I_1 as compared to integrating I_2, thats probably why.

Sorry for the mistake. Oh ok I though that but I though it should work so I discarded that thought.

Davo_01 · Apr 10, 2014

Re: MX2 Integration Marathon

hit patel said:
Sorry for the mistake. Oh ok I though that but I though it should work so I discarded that thought.

Yeh the formula should work even for n>=3 but to end up with and integral we can easily evaluate, we would want to sub a number in the form n=3k+2

Sy123 · Apr 10, 2014

Re: MX2 Integration Marathon

$\int_2^6 \ln \left(\frac{-1+\sqrt{1+4x}}{2} \right) \ dx$

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_2^6 \ln \left(\frac{-1+\sqrt{1+4x}}{2} \right) \ dx$

LET M BE the INTEGRAL PROVIDED.
Let u^2= 1+4x (u.du)/2 = dx when x=6, u= 5 when x= 2 u= 3
Integral _5_3((ln(-1+u) -ln2) u.du/2)
--------------------------------------------------------------
Let integral _5 _3 (u.ln(u-1))= I
For I,
Let y= ln (u-1) dy=1/(u-1)
Let dz= u z= (u^2)/2
therefore I= [((u^2)/2)ln(u-1)]_5_3 - integral ((u^2)/2(u-1))_5 _3
= (41/2)ln(2) - integral (u + u/(u-1))_5_3
= 20ln(2) -5
-----------------------------------------------
M= (1/2) (20ln(2)-5) - 4ln(2) = 6 ln(2) - 5/2

____________________________________________________________________
Sorry for the messy work. Possibly and probably some mistakes due to bracketing problems that I faced.

Sy123 · Apr 10, 2014

Re: MX2 Integration Marathon

hit patel said:
LET M BE the INTEGRAL PROVIDED.
Let u^2= 1+4x (u.du)/2 = dx when x=6, u= 5 when x= 2 u= 3
Integral _5_3((ln(-1+u) -ln2) u.du/2)
--------------------------------------------------------------
Let integral _5 _3 (u.ln(u-1))= I
For I,
Let y= ln (u-1) dy=1/(u-1)
Let dz= u z= (u^2)/2
therefore I= [((u^2)/2)ln(u-1)]_5_3 - integral ((u^2)/2(u-1))_5 _3
= (41/2)ln(2) - integral (u - u/(u-1))_5_3
= (41/2)ln(2) - 6 + ln (3/5)
-----------------------------------------------
M= (41/2)ln(2) -6 + ln (3/5) -16 ln (2) = (9/2)ln(2) + ln(3/5) -6

____________________________________________________________________
Sorry for the messy work. Possibly and probably some mistakes due to bracketing problems that I faced.

I think you have made a mistake

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

hit patel said:
LET M BE the INTEGRAL PROVIDED.
Let u^2= 1+4x (u.du)/2 = dx when x=6, u= 5 when x= 2 u= 3
Integral _5_3((ln(-1+u) -ln2) u.du/2)
--------------------------------------------------------------
Let integral _5 _3 (u.ln(u-1))= I
For I,
Let y= ln (u-1) dy=1/(u-1)
Let dz= u z= (u^2)/2
therefore I= [((u^2)/2)ln(u-1)]_5_3 - integral ((u^2)/2(u-1))_5 _3
= (41/2)ln(2) - integral (u - u/(u-1))_5_3
= (41/2)ln(2) - 6 + ln (3/5)
-----------------------------------------------
M= (41/2)ln(2) -6 + ln (3/5) -16 ln (2) = (9/2)ln(2) + ln(3/5) -6

____________________________________________________________________
Sorry for the messy work. Possibly and probably some mistakes due to bracketing problems that I faced.

Used curve sketching to draw the graph found an approximate area to find integral. Answer way off and got a negative answer that doesnt correspond to the answer I got.

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
I think you have made a mistake

Think i fixed it.

Davo_01 · Apr 10, 2014

Re: MX2 Integration Marathon

6ln2-5/2? Just use u=ln(...) And hence x=e^u+e^2u then IBP

Sy123 · Apr 10, 2014

Re: MX2 Integration Marathon

hit patel said:
Think i fixed it.

Yep that is correct.

The method I had in mind was recognising that the argument inside the logarithm looks like a solution to a quadratic equation, namely

$z^2+z - x = 0$

So it makes sense then to just make the substitution

$x = u^2+u$

$dx = (2u+1) \ du$

$\frac{-1+ \sqrt{1+4x}}{2} = u$

Then simple by parts from there

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Sy123 said:
Yep that is correct.

The method I had in mind was recognising that the argument inside the logarithm looks like a solution to a quadratic equation, namely

$z^2+z - x = 0$

So it makes sense then to just make the substitution

$x = u^2+u$

$dx = (2u+1) \ du$

$\frac{-1+ \sqrt{1+4x}}{2} = u$

Then simple by parts from there

Oh ok. I didnt see that. I have never seen anyone use that. That a really great method. Cant rep. I would If i could.

hit patel · Apr 10, 2014

Re: MX2 Integration Marathon

Davo_01 said:
6ln2-5/2? Just use u=ln(...) And hence x=e^u+e^2u then IBP

Yep thanks.

seanieg89 · Apr 13, 2014

Re: MX2 Integration Marathon

seanieg89 said:
In this question we compute the Dirichlet integral using high school methods.

$1. Show that: \\ \\$\int_0^{\pi/2} \frac{\sin((2n+1)x)}{\sin(x)}\, dx = \frac{\pi}{2}.$\\ \\ 2. Let: \\ \\$I_n:=\int_0^{\frac{(2n+1)\pi}{2}}\frac{\sin(x)}{x}\, dx$\\ and show that $D_n=\frac{\pi}{2}-I_n \rightarrow 0$ by using 1.\\ \\ 3. Hence, given that the limit: $I:=\lim_{T\rightarrow \infty}\left(\int_0^T \frac{\sin(x)}{x}\, dx\right) $ exists, find it.$

Note: n ranges over the non-negative integers.

.

hit patel · Apr 13, 2014

Re: MX2 Integration Marathon

seanieg89 said:
.

1) sin (2n+1)x = sin (2n-1 +2)x
= sin(2n-1)x cos(2x) - cos(2n-1)xsin(2x)
= sin (2n-1)x (1-2sin^(2) x) - cos (2n-1)x(2cosxsinx)
Expanding we get:
sin (2n-1)x - 2sin(2n-1)xsin^(2)x + 2cos(2n-1)xsinxcosx
Now sin(2n+1)x/sinx = sin (2n-1)x / sin(x) - 2sin (2n-1)xsinx + 2cos(2n-1)xcosx
But this consists of cos(a+b)
Therefore = sin (2n-1)x/sin (x) -2 cos(2nx)
Integral _pi/2_0 sin (2(n-1)+1)x/sin(x) -2cos(2nx) dx

= U_(n-1) + 2 [ sin(2nx)/2n] _ pi/2_0
= U_(n-1)

huh? now? Looks like I did something wrong. how can integral of u_n be equal to u_(n-1) hm? and how do I continue .
isnt 2n-1 always odd in my answer therefore i get 1 as answer/

And also any other quicker ways. This way took me a long time to think of because first time I did it I expanded straight away. Then thought about the ways I could integrate and saw that the only way I could do this and get rid of the sinx at bottom was by using double angle formula so yeh I had to go through a long process. Any other shorter methods?

RealiseNothing · Apr 13, 2014

Re: MX2 Integration Marathon

hit patel said:
1) sin (2n+1)x = sin (2n-1 +2)x
= sin(2n-1)x cos(2x) - cos(2n-1)xsin(2x)
= sin (2n-1)x (1-2sin^(2) x) - cos (2n-1)x(2cosxsinx)
Expanding we get:
sin (2n-1)x - 2sin(2n-1)xsin^(2)x + 2cos(2n-1)xsinxcosx
Now sin(2n+1)x/sinx = sin (2n-1)x / sin(x) - 2sin (2n-1)xsinx + 2cos(2n-1)xcosx
But this consists of cos(a+b)
Therefore = sin (2n-1)x/sin (x) -2 cos(2nx)
Integral _pi/2_0 sin (2(n-1)+1)x/sin(x) -2cos(2nx) dx

= U_(n-1) + 2 [ sin(2nx)/2n] _ pi/2_0
= U_(n-1)

huh? now? Looks like I did something wrong. how can integral of u_n be equal to u_(n-1) hm? and how do I continue .
isnt 2n-1 always odd in my answer therefore i get 1 as answer/

$u_n=u_{n-1}=u_{n-2}=...=u_2=u_1=u_0$

$u_0 = \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sin(x)} ~dx$

$u_0 = \int_{0}^{\frac{\pi}{2}} 1~dx$

$\frac{\pi}{2}$

hit patel · Apr 13, 2014

Re: MX2 Integration Marathon

RealiseNothing said:
$u_n=u_{n-1}=u_{n-2}=...=u_2=u_1=u_0$

$u_0 = \int_{0}^{\frac{\pi}{2}} \frac{\sin(x)}{\sin(x)} ~dx$

$u_0 = \int_{0}^{\frac{\pi}{2}} 1~dx$

$\frac{\pi}{2}$

Hahahah what so I actually had the answer the whole time and I went through all that. Hahaha I used that fact you just showed in the answer I wrote. Look at the bottom of my answer. What was I thinking? But shouldnt I still get the same answer?

juantheron · Apr 16, 2014

Re: MX2 Integration Marathon

$How\ can\; we\; prove \; \displaystyle \int_{0}^{\frac{\pi}{4}}\tan^{-1}\left(\frac{\sqrt{2}\cos3 \phi}{\left(2\cos 2 \phi+ 3\right)\sqrt{\cos 2 \phi}}\right)d\phi = 0$

Sy123 · Apr 18, 2014

Re: MX2 Integration Marathon

$\int \frac{dx}{(x+\alpha)^{\frac{n+1}{n}} (x+\beta)^{\frac{n-1}{n}}}$

HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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