HSC 2015 MX2 Integration Marathon (archive) (3 Viewers)

swagswagyoloswag · Apr 10, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Here's another recurrence question:

$$ Let $ I_{n} = \int_{0}^{\frac{\pi}{2}} \sin^{n}x \cos^{2} x dx$

$$ Show that $ I_{n} = \frac{n-1}{n+2} \Cdot I_{n-2}$

I'm thinking integration by parts by letting u = sin^(n-1)(x) and dv = sinxcos^2 (x)then you'll approach a step with integration of sin^(n-2)cos^4 (x) and just change cos^4(x) to (1-sin^2 (x))(cos^(2)(x)

Sy123 · Apr 11, 2015

Re: MX2 2015 Integration Marathon

$\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$

Ekman · Apr 11, 2015

Re: MX2 2015 Integration Marathon

Sy123 said:
$\int \frac{dx}{\sqrt{x} + \sqrt[3]{x}}$

$$Let $u^6=x \therefore \int \frac{dx}{\sqrt{x} + \sqrt[3]{x}} = \int \frac{6u^3}{u+1} du$

$=6\int u^2 -u+1-\frac{1}{u+1}du = 2u^3 -3u^2+6u-6ln(u+1) +c$

$\therefore 2\sqrt{x} -3\sqrt[3]{x} + 6\sqrt[6]{x} - 6ln(\sqrt[6]{x}+1) +c$

VBN2470 · Apr 11, 2015

Re: MX2 2015 Integration Marathon

NEW Q:

$If $I_n = \int_{0}^{1} x^{n}\sqrt{1-x}\text{ d}x$ show that $I_n = \frac{2n}{2n+3}I_{n-1}$ for $n\,\geq\,1$. \\ Hence, show that $I_n = \frac{n!(n+1)!}{(2n+3)!}4^{n+1}$$

seanieg89 · Apr 11, 2015

Re: MX2 2015 Integration Marathon

seanieg89 said:
Here is a question that concerns inequalities arising from integration. (It is harder conceptually than most questions in this thread, but easier than a lot of them in terms of how technically demanding the required manipulations are.)

$Let $\alpha>0$ be a positive parameter.\\ Show that there exist positive constants $C_1,C_2$ such that: \\ \\ $C_1||z|-1|^{-\alpha}<\int_0^{2\pi} |z-\textrm{cis}(\theta)|^{-\alpha} \, d\theta < C_2||z|-1|^{-\alpha}$ for $|z|\neq 1$.$

(This integral clearly blows up as $z$ approaches the unit circle. The point of the question is to quantify how quickly this happens, which is generally a useful thing to know.)

Bump. Note that the upper bound has already been proved by Integrand above.

Drsoccerball · Apr 11, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW Q:

$If $I_n = \int_{0}^{1} x^{n}\sqrt{1-x}\text{ d}x$ show that $I_n = \frac{2n}{2n+3}I_{n-1}$ for $n\,\geq\,1$. \\ Hence, show that $I_n = \frac{n!(n+1)!}{(2n+3)!}4^{n+1}$$

I was able to do all of it but i got $I_n= \frac{n!(n+1)!4^n}{(2n+3)!}$ where does the extra 4 come from ?

VBN2470 · Apr 11, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I was able to do all of it but i got $I_n= \frac{n!(n+1)!4^n}{(2n+3)!}$ where does the extra 4 come from ?

I can assure you it is not a typo this time

VBN2470 · Apr 11, 2015

Re: MX2 2015 Integration Marathon

NEW Q:

$(i)\, Suppose that $x$ is a positive real number and that $n$ is a positive integer. Show that $\frac{1}{1+x^n}\,<\,1$. \\ (ii)\, Let $I_n = \int_{0}^{1} \frac{1}{1+x^n}\text{ d}x$ where $n$ is a positive integer and $n\,\geq\,2$. Show that $\frac{\pi}{4}\,\leq\,I_n\,<\,1$$

Ekman · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Interesting Question:

$\int e^x \sin{2x} \cos{x} dx$

swagswagyoloswag · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Interesting Question:

$\int e^x \sin{2x} \cos{x} dx$

Change sin2xcosx into 1/2 (sin3x +sinx) and expand e^x, split the integral and use integration by parts for each. There might be a faster way

swagswagyoloswag · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Interesting Question:

$\int e^x \sin{2x} \cos{x} dx$

Or I'm thinking change sin2x into 2sinxcosx which then your integral becomes e^x sin(x)cos^2(x) then use IBP by letting u = e^x and dv = the rest. You then get 1/3 e^x cos^3(x) - 1/3 int e^x cos^3(x). Let the new integral be J and so J = int e^x cosx - e^x cosxsin^2(x). Then repeat IBP. lol don't have pen or paper here so can't write it down

Drsoccerball · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Interesting Question:

$\int e^x \sin{2x} \cos{x} dx$

I think theres an easier way then integration by parts

swagswagyoloswag · Apr 12, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW Q:

$(i)\, Suppose that $x$ is a positive real number and that $n$ is a positive integer. Show that $\frac{1}{1+x^n}\,<\,1$. \\ (ii)\, Let $I_n = \int_{0}^{1} \frac{1}{1+x^n}\text{ d}x$ where $n$ is a positive integer and $n\,\geq\,2$. Show that $\frac{\pi}{4}\,\leq\,I_n\,<\,1$$

$$ (i) $ x > 0 \\ x^{n} > 0 \\ 1+x^{n} > 1 \\ \frac{1}{1+x^{n}} < 1 \\ $ (ii) $ n\, \geq\,2 $ \\ 1 + x^{n} < 1 + x^{n+1} \\ \frac{1}{1+x^{n}} > \frac{1}{1+x^{n+1}} \\ $ For the interval $ 1 \geq\,x \geq\,0 \\ x^{2} \geq\,x^{n} \\ \frac{1}{1+x^{n}} \geq\,\frac{1}{1+x^{2}} \\ \int_{0}^{1} \frac{1}{1+x^{n}} \geq\,\int_{0}^{1} \frac{1}{1+x^{2}} \\ I_n \geq\, [tan^{-1}(x)]_{0}^{1} \\ I_n \geq\,\frac{\pi}{4} \\ $ From (i), \frac{1}{1+x^{n}} < 1 $ \\ \int_{0}^{1} \frac{1}{1+x^{n}} < \int_{0}^{1} 1 dx \\ I_n < 1 \\ \therefore \frac{\pi}{4} \leq\, I_n < 1$

VBN2470 · Apr 12, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Find $\int_{1}^{\infty} \frac{\text{ d}x}{x{\sqrt{x^2+2x-1}}}$$

integral95 · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
I was able to do all of it but i got $I_n= \frac{n!(n+1)!4^n}{(2n+3)!}$ where does the extra 4 come from ?

$I_0 = 2/3$

when you multiply top and bottom by (2n+2)(2n)....(4)(2) you realise there are n+1 2s to factorise

Drsoccerball · Apr 12, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
$$ (i) $ x > 0 \\ x^{n} > 0 \\ 1+x^{n} > 1 \\ \frac{1}{1+x^{n}} < 1 \\ $ (ii) $ n\, \geq\,2 $ \\ 1 + x^{n} < 1 + x^{n+1} \\ \frac{1}{1+x^{n}} > \frac{1}{1+x^{n+1}} \\ $ For the interval $ 1 \geq\,x \geq\,0 \\ x^{2} \geq\,x^{n} \\ \frac{1}{1+x^{n}} \geq\,\frac{1}{1+x^{2}} \\ \int_{0}^{1} \frac{1}{1+x^{n}} \geq\,\int_{0}^{1} \frac{1}{1+x^{2}} \\ I_n \geq\, [tan^{-1}(x)]_{0}^{1} \\ I_n \geq\,\frac{\pi}{4} \\ $ From (i), \frac{1}{1+x^{n}} < 1 $ \\ \int_{0}^{1} \frac{1}{1+x^{n}} < \int_{0}^{1} 1 dx \\ I_n < 1 \\ \therefore \frac{\pi}{4} \leq\, I_n < 1$

See how your Latex is squashed for each new line what i do is i start a new *tex* but by leaving lines betwwen each one

swagswagyoloswag · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
See how your Latex is squashed for each new line what i do is i start a new *tex* but by leaving lines betwwen each one

$so$
$\int{a}^{b}$
$like$
$\frac{c}{d}$
$this$
Thanks!

Ekman · Apr 12, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int_{0}^{\frac{\pi}{4}} \frac{\sec(x)}{2+\sin^2(x)} dx$

VBN2470 · Apr 12, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{1}^{\infty} \frac{\text{ d}x}{x{\sqrt{x^2+2x-1}}}$$

BUMP

Ekman · Apr 12, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{1}^{\infty} \frac{\text{ d}x}{x{\sqrt{x^2+2x-1}}}$$

Im sure there is a more elegant method:

$\int_{1}^{\infty} \frac{\text{ d}x}{x{\sqrt{x^2+2x-1}}}$ $ = \int_{1}^{\infty} \frac{\text{ d}x}{x{\sqrt{(x+1)^2-2}}}$

$$ Let $ x+1=\sqrt{2}\sec(\theta)$

$$ Hence: $ \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{\sec(\theta)}{\sqrt{2}\sec(\theta) -1} d\theta$

$= \int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \frac{1}{\sqrt{2} -\cos(\theta)} d\theta = \left[2\tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)(\sqrt{2}+1)\right)\right]_{\frac{\pi}{4}}^{\frac{\pi}{2}} $ Via t-method$$

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