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

hit patel · May 17, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\pi/2} \frac{\sin^2(n\theta)}{\sin^2 \theta} \ d\theta$

int(f(x))=int((f(a-x))?

Sy123 · May 17, 2014

Re: MX2 Integration Marathon

hit patel said:
int(f(x))=int((f(a-x))?

That will probably not work, this is beyond HSC difficulty though (but within the scope of the syllabus)

The answer is $n\frac{\pi}{2}$ , it uses recursion in the solution, I can post my one on request

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$\int_0^{\pi /4}\frac{\tan x}{1 + \sin x} \ dx$

Davo_01 · May 17, 2014

Re: MX2 Integration Marathon

Sy123 said:
That will probably not work, this is beyond HSC difficulty though (but within the scope of the syllabus)

The answer is $n\frac{\pi}{2}$ , it uses recursion in the solution, I can post my one on request

-----------------

$\int_0^{\pi /4}\frac{\tan x}{1 + \sin x} \ dx$

Had a feeling there would be recursion, took me a while to find it:

$\dfrac{\sin^2{nx}}{\sin^2{x}}=\dfrac{\sin^2{(n-1)x}}{\sin^2{x}}+\dfrac{\sin{(2n-1)}}{\sin{x}}$

$$Let$ \; I_n=\int_0^{\frac{\pi}{2}}\dfrac{\sin^2{nx}}{\sin^2{x}}dx=I_{n-1}+\int_0^{\frac{\pi}{2}}\dfrac{\sin(2n-1)}{\sin{x}}dx$

$\dfrac{\sin{(2n-1)x}}{\sin{x}}=\dfrac{\sin(2n-3)x}{\sin{x}}+2\cos(2(n-1)x)$

$$Note: $\;\int_0^{\frac{\pi}{2}}\cos(2kx)dx=0$

$\int_0^{\frac{\pi}{2}}\dfrac{\sin{(2n-1)x}}{\sin{x}}dx$ $= \int_0^{\frac{\pi}{2}}\dfrac{\sin{(2n-3)x}}{\sin{x}}dx\;=\int_0^{\frac{\pi}{2}}\dfrac{\sin{x}}{\sin{x}}dx=\frac{\pi}{2}$

$\therefore I_n=I_{n-1}+\frac{\pi}{2}$ $\;=\frac{n\pi}{2}$

Ikki · May 17, 2014

Re: MX2 Integration Marathon

Sy123 said:
That will probably not work, this is beyond HSC difficulty though (but within the scope of the syllabus)

The answer is $n\frac{\pi}{2}$ , it uses recursion in the solution, I can post my one on request

-----------------

$\int_0^{\pi /4}\frac{\tan x}{1 + \sin x} \ dx$

Step back, davo's sol^n is like crazy. Sy did u have to mess around with identities as well or something else?

Alright, while i'm here...
Let t=tan(x/2)

$\int_0^{tan(pi/8)}\frac{\frac{2t}{1-t^2}}{1+\frac{2t}{1+t^2}} * \frac{2}{1+t^2} \ dt$

$=2\int_0^{tan(pi/8)}\frac{\frac{2t}{1-t^2}}{1+t^2+2t} \ dt$

$=2\int_0^{tan(pi/8)}\frac{2t}{(1+t)^3(1-t)} \ dt$

Now casually using partial fractions...

$=2\int_0^{tan(pi/8)}\frac{-1/4}{t-1}-\frac{1}{(1+t)^3}+\frac{1/2}{(1+t)^2}+\frac{1/4}{1+t} \ dt$

At which point I basically ceebs typing the rest up, but should be integratable now easily.

RealiseNothing · May 17, 2014

Re: MX2 Integration Marathon

Not sure if I've posted this question before:

$$Define$~I_n = \int_0^1 \frac{(1-x)^n}{1+x}~dx$

$$Find$~\sum_{k=1}^n \frac{2}{k}I_k$

If you are HSC you may need to know a bit of:

http://en.wikipedia.org/wiki/Taylor_series

Sy123 · May 17, 2014

Re: MX2 Integration Marathon

Ikki said:
Step back, davo's sol^n is like crazy. Sy did u have to mess around with identities as well or something else?

Alright, while i'm here...
Let t=tan(x/2)

$\int_0^{tan(pi/8)}\frac{\frac{2t}{1-t^2}}{1+\frac{2t}{1+t^2}} * \frac{2}{1+t^2} \ dt$

$=2\int_0^{tan(pi/8)}\frac{\frac{2t}{1-t^2}}{1+t^2+2t} \ dt$

$=2\int_0^{tan(pi/8)}\frac{2t}{(1+t)^3(1-t)} \ dt$

Now casually using partial fractions...

$=2\int_0^{tan(pi/8)}\frac{-1/4}{t-1}-\frac{1}{(1+t)^3}+\frac{1/2}{(1+t)^2}+\frac{1/4}{1+t} \ dt$

At which point I basically ceebs typing the rest up, but should be integratable now easily.

Yea his solution was pretty much exact same as mine

As for your solution to this, it is quite long and you will need to evaluate $\tan \frac{\pi}{8}$

Here is my solution:

$\int_0^{\pi/4} \frac{\tan x }{1 + \sin x} \ dx = \int_0^{\pi}{4} \frac{\tan x (1-\sin x)}{\cos^2x} \ dx \\ \\ = \int_0^{\pi/4} \tan x \sec^2 x - \tan^2 x \sec x \ dx = \frac{1}{2} \tan^2 x |_0^{\pi/4} - \underbrace{\int_0^{\pi/4} \tan^2x \sec x \ dx}_{I} \\ \\ I = \int_0^{\pi}{4} \sec^3 x - \sec x \ dx = \underbrace{\int_0^{\pi /4} \sec^3 x \ dx}_J - \ln (\sqrt{2} + 1) \\ \\ J = \int_0^{\pi/4} \sec^3 x \dx \\ u = \sec x \Rightarrow \ du = \sec x \tan x \\ dv = \sec^2x \ dx \Rightarrow \ v = \tan x \\ \\ \therefore \ J = \sqrt{2} - I = \sqrt{2} - (J - \ln(\sqrt{2} +1)) \\ \\ \therefore \ J = \frac{1}{\sqrt{2}} + \frac{1}{2}\ln(\sqrt{2} + 1) \\ \\ \therefore \ \int_0^{\pi/4} \frac{\tan x}{1+\sin x} \ dx = \frac{1}{2} - \frac{1}{\sqrt{2}} + \ln \sqrt{\sqrt{2} + 1}$

Ikki · May 18, 2014

Re: MX2 Integration Marathon

Ah k nice

Question: How would you determine value for tan (pi/8)?

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Nvm. Figured it out. You can just split it into tan(pi/4-pi/8) and solve the quadratic.

Ikki · May 20, 2014

Re: MX2 Integration Marathon

New Question!!!

$\int \frac{3\sqrt2}{18x^4+2}\cdot x \cdot arccot(3x^2)\cdot dx$

Drongoski · May 20, 2014

Re: MX2 Integration Marathon

$I = \frac {\sqrt 2}{4} \int \frac {1}{1 + (3x^2)^2} \cdot arccot(\3x^2) \cdot d(3x^2)$

$= - \frac {\sqrt 2}{4} \int arccot(3x^2) d(arccot(3x^2))$

$= -\frac {\sqrt 2}{8} \left [ arccot(3x^2) \right ]^2 + C$

Can someone pose a new question, on my behalf, please.

Sy123 · May 21, 2014

Re: MX2 Integration Marathon

$\int_0^{\pi /2} \frac{1}{1+\sin^2x} \ dx$

seanieg89 · May 21, 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.

.

dunjaaa · May 21, 2014

Re: MX2 Integration Marathon

Screen shot 2014-05-21 at 8.38.28 AM.png

Ikki · May 21, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30472

Just a minor mistake... From your working:

$\frac{1}{\sqrt2}\int_0^{\pi /2} \frac{\sqrt{2}sec^2x}{1+2tan^2x} \ dx$

Now if you observe closely, the integral above is actually the tan inverse of the tan function.

Thus,

$= \frac{1}{\sqrt2} \cdot [x]_{0}^{pi/2}$

$= \frac{\pi}{2\sqrt{2}}$

Ikki · May 21, 2014

Re: MX2 Integration Marathon

Drongoski said:
$I = \frac {\sqrt 2}{4} \int \frac {1}{1 + (3x^2)^2} \cdot arccot(\3x^2) \cdot d(3x^2)$

$= \frac {\sqrt 2}{4} \int arccot(3x^2) d(arccot(3x^2))$

$= \frac {\sqrt 2}{8} \left [ arccot(3x^2) \right ]^2 + C$

Can someone pose a new question, on my behalf, please.

Did not think of that, if you let u=arccot(3x^2) then you end up with a ln? Are both correct?

Also, in regards to the working, derivative of arccotx=negative * 1/1+x^2. So i believe there is a negative missing.

Drongoski · May 21, 2014

Re: MX2 Integration Marathon

Ikki said:
Also, in regards to the working, derivative of arccotx=negative * 1/1+x^2. So i believe there is a negative missing.

You are correct. There should be a negative. When I tentatively started doing it, I thought there ought to be a negative. I then flipped thru 2 text books for the derivative of arccot x - looking out for the '-' sign. Both times I did not see the '-'sign. After seeing your feedback, I rechecked both and there was the '-'sign. How I missed it both times yesterday baffles me. It must mean my glasses are not working as well as I thought.

By the way, if you apply the Chain Rule to [arccot ( (3x^2)]^2 you will get a constant multiple of the integrand. That is you will get:

$2(cot(3x^2)) \times \frac {-1}{1 + (3x^2)^2} \times 6x$

Ikki · May 21, 2014

Re: MX2 Integration Marathon

Epic.

So, If i let u=arctan(3x^2) right, i get sqrt(2)/4 * ln(arctan(3x^2)) which if you derive also get a constant multiple of the integrand, so i assume both are correct.

Drongoski · May 21, 2014

Re: MX2 Integration Marathon

Ikki said:
Epic.

So, If i let u=arctan(3x^2) right, i get sqrt(2)/4 * ln(arctan(3x^2)) which if you derive also get a constant multiple of the integrand, so i assume both are correct.

Where does this 'ln' come from ???

dunjaaa · May 21, 2014

Re: MX2 Integration Marathon

Woops typo, was meant to be arctan(sqrt(2)tan(x)) instead of arctan(1+sqrt(2)tan(x)), typed it on the bus this morning

Ikki · May 21, 2014

Re: MX2 Integration Marathon

Drongoski said:
Where does this 'ln' come from ???

Whoops I took cot inverse x= 1/tan inverse x

abecina · May 21, 2014

Re: MX2 Integration Marathon

$\int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}$
(without using complex numbers)

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