HSC 2015 MX2 Integration Marathon (archive) (1 Viewer)

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$\int{\frac{1}{\sqrt{x}-1}}dx$

$\int{\frac{1}{\sqrt{x}-1}}dx = \int \frac{\sqrt{x}+1}{x-1}$ Rationalized the denominator.

$$ Split the integrand to get: $ \int \frac{\sqrt{x}}{x-1}dx + \int \frac{1}{x-1} dx$

$\int \frac{\sqrt{x}}{x-1} dx = \int \frac{2u^2}{u^2-1}du$ (Through x=u^2 substitution)

$\therefore \int \frac{2u^2}{u^2-1}du = 2u + ln(|\frac{u-1}{u+1}|) = 2\sqrt{x} + ln(|\frac{\sqrt{x}-1}{\sqrt{x}+1}|)$

$\int \frac{1}{x-1} = ln(|x-1|)$

$\therefore \int{\frac{1}{\sqrt{x}-1}}dx = 2\sqrt{x} + ln(|\frac{\sqrt{x}-1}{\sqrt{x}+1}|) + ln(|x-1|) +c$

Edit: I just realized now:
$=2\sqrt{x} + 2ln(|\sqrt{x}-1|) + c$

swagswagyoloswag · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int{\frac{1}{\sqrt{x}-1}}dx = \int \frac{\sqrt{x}+1}{x-1}$ Rationalized the denominator.

$$ Split the integrand to get: $ \int \frac{\sqrt{x}}{x-1}dx + \int \frac{1}{x-1} dx$

$\int \frac{\sqrt{x}}{x-1} dx = \int \frac{2u^2}{u^2-1}du$ (Through x=u^2 substitution)

$\therefore \int \frac{2u^2}{u^2-1}du = 2u + ln(|\frac{u-1}{u+1}|) = 2\sqrt{x} + ln(|\frac{\sqrt{x}-1}{\sqrt{x}+1}|)$

$\int \frac{1}{x-1} = ln(|x-1|)$

$\therefore \int{\frac{1}{\sqrt{x}-1}}dx = 2\sqrt{x} + ln(|\frac{\sqrt{x}-1}{\sqrt{x}+1}|) + ln(|x-1|) +c$

Edit: I just realized now:
$=2\sqrt{x} + 2ln(|\sqrt{x}-1|) + c$

Couldn't you have just let u^2 = x as the first step?

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$\int{\frac{1}{\sqrt{x}-1}}dx = \int \frac{\sqrt{x}+1}{x-1}$ Rationalized the denominator.

$$ Split the integrand to get: $ \int \frac{\sqrt{x}}{x-1}dx + \int \frac{1}{x-1} dx$

$\int \frac{\sqrt{x}}{x-1} dx = \int \frac{2u^2}{u^2-1}du$ (Through x=u^2 substitution)

$\therefore \int \frac{2u^2}{u^2-1}du = 2u + ln(|\frac{u-1}{u+1}|) = 2\sqrt{x} + ln(|\frac{\sqrt{x}-1}{\sqrt{x}+1}|)$

$\int \frac{1}{x-1} = ln(|x-1|)$

$\therefore \int{\frac{1}{\sqrt{x}-1}}dx = 2\sqrt{x} + ln(|\frac{\sqrt{x}-1}{\sqrt{x}+1}|) + ln(|x-1|) +c$

Edit: I just realized now:
$=2\sqrt{x} + 2ln(|\sqrt{x}-1|) + c$

$let u^2 =x$

$2u du =dx$

$\therefore $ integral becomes $ \int{2 + \frac{2}{u-1}}$

$2u -2ln|u-1|$

$= 2\sqrt{x} -2ln|\sqrt{x}-1| +C$

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
Couldn't you have just let u^2 = x as the first step?

Or that, when I saw the question, rationalizing the denominator was the first thing that came to my mind, but of course your substitution is quicker.

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

can we get some recurrence questions up in this b***h

SpiralFlex · Apr 6, 2015

Re: MX2 2015 Integration Marathon

$\textup{Let} \;a < b. \;\textup{Show that,}$

$\int_{p}^{q}\sqrt{\frac{x-a}{b-x}}\;dx = \frac{(b-a)(\pi+3\sqrt{3}-6)}{12}$

$\textup{Where} \;p = \frac{3a+b}{4} \;\textup{and}\; q = \frac{a+b}{2}$

swagswagyoloswag · Apr 6, 2015

Re: MX2 2015 Integration Marathon

SpiralFlex said:
$\textup{Let} \;a < b. \;\textup{Show that,}$

$\int_{p}^{q}\sqrt{\frac{x-a}{b-x}}\;dx = \frac{(b-a)(\pi+3\sqrt{3}-6)}{12}$

$\textup{Where} \;p = \frac{3a+b}{4} \;\textup{and}\; q = \frac{a+b}{2}$

Is there a lot of algebra? At least I got the (b-a) part lol

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
Is there a lot of algebra? At least I got the (b-a) part lol

Im thinking (because i love triangles) making two seperate triangle and finding expresions for the angles and then just using substitution and bam answer but ceebs

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

SpiralFlex said:
$\textup{Let} \;a < b. \;\textup{Show that,}$

$\int_{p}^{q}\sqrt{\frac{x-a}{b-x}}\;dx = \frac{(b-a)(\pi+3\sqrt{3}-6)}{12}$

$\textup{Where} \;p = \frac{3a+b}{4} \;\textup{and}\; q = \frac{a+b}{2}$

Are you sure that the answer is:
$\frac{(b-a)(\pi+3\sqrt{3}-6)}{12}$

Because I got:
$\frac{(b-a)(2\pi+3\sqrt{3}-6)}{12}$

The integrated version for me was:
$(b-a) \left [\frac{\sqrt{b-x}\cdot\sqrt{x-a}}{b-a} + \sin^{-1}\left(\frac{\sqrt{b-x}}{\sqrt{b-a}}\right)\right]_{q}^{p}$

swagswagyoloswag · Apr 6, 2015

Re: MX2 2015 Integration Marathon

I got a final answer of $(\frac{-3 \pi}{2} ) ( \frac{b-a}{2})$

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
I got a final answer of $(\frac{-3 \pi}{2} ) ( \frac{b-a}{2})$

I let u^2=b-x and had as a result:
$2\int -\sqrt{(b-a)-u^2}du$

Via triangles I got to the final integral:
$2(b-a)\int cos^2(\theta) d\theta $ Where $\theta=\sin^{-1}\left(\frac{u}{\sqrt{b-a}}\right)$

After integrating, I expanded to get the final integrated version in my previous post

swagswagyoloswag · Apr 6, 2015

Re: MX2 2015 Integration Marathon

found an algebra mistake and got [(b-a)(3sqrt3 - 6 - 3pi)]/12 lol

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
found an algebra mistake and got [(b-a)(12-pi - 3sqrt3)]/12 lol

What was your final integrated indefinite answer?

swagswagyoloswag · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
What was your final integrated indefinite answer?

$\frac{1}{2}(b-a)sin^{-1}(\frac{4x-(2a+2b)}{a-b}) - \sqrt{\frac{(a-b)^2 }{4} - (x - \frac{a+b}{2} )^2 $ from p to q $$
can you sub it in for me? lol I probably made some mistake again
http://m.imgur.com/xu1MKbH

Ekman · Apr 6, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Are you sure that the answer is:
$\frac{(b-a)(\pi+3\sqrt{3}-6)}{12}$

Because I got:
$\frac{(b-a)(2\pi+3\sqrt{3}-6)}{12}$

The integrated version for me was:
$(b-a) \left [\frac{\sqrt{b-x}\cdot\sqrt{x-a}}{b-a} + \sin^{-1}\left(\frac{\sqrt{b-x}}{\sqrt{b-a}}\right)\right]_{q}^{p}$

Sorry guys, I just found my mistake.

$\frac{(b-a)(\pi+3\sqrt{3}-6)}{12}$ Is correct. When I subbed in q, I forgot to square root it inside the arcsin term, so I got pi/6 rather than pi/4.

VBN2470 · Apr 6, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Find $\int_{0}^{\infty} \frac{|x-1|}{(x+1)(x^2+1)} \text{ d}x$$

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{0}^{\infty} \frac{|x-1|}{(x+1)(x^2+1)} \text{ d}x$$

for all values of x (repeat process and get the same answer)
$\int_0^{\infty}{ \frac{|x-1|}{(x+1)(x^2+1)}$

$Partial fractions gives you$

$\frac{-1}{x+1} + \frac{x}{x^2 +1}$

$(ln (\frac{\sqrt{x^2 +1}}{x+1}) $ With integral values of 0 to infinity $$

$0$

swagswagyoloswag · Apr 6, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Find $\int_{0}^{\infty} \frac{|x-1|}{(x+1)(x^2+1)} \text{ d}x$$

= $\int_{0}^{1} \frac{1-x}{(x+1)(x^2 + 1)} + \int_{1}^{\infty} \frac{x-1}{(x+1)(x^2 + 1)} \\ = [\frac{-1}{2} ln(x^{2}+1) - ln(x+1) ]_{0}^{1} + [ln(\frac{ \sqrt{x^2 + 1}}{x+1})]_{1}^{\infty} \\ = \frac{1}{2}ln2 - ln(\frac{\sqrt{2}}{2}}) = ln2$

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

swagswagyoloswag said:
= $\int_{0}^{1} \frac{1-x}{(x+1)(x^2 + 1)} + \int_{1}^{\infty} \frac{x-1}{(x+1)(x^2 + 1)} \\ = [\frac{-1}{2} ln(x^{2}+1) - ln(x+1) ]_{0}^{1} + [\frac{1}{2} ln(\frac{x^2 + 1}{x+1})]_{1}^{infty} \\ = \frac{1}{2}ln2 - ln(\frac{\sqrt{2}}{2}}) = ln2$

ahhh my bad

Drsoccerball · Apr 6, 2015

Re: MX2 2015 Integration Marathon

$\int_0^{\frac{\pi}{2}}{\frac{dx}{a^2sin^2x +b^2cos^2x}}$

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