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

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
change the tanx tan2xtan3x into tan3x -tan2x - tanx by considering the expansion of tan (2x+x). then pretty straightforward from there

Hsc 2016?

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

New question

$\int_0^2{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}dx$

InteGrand · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
New question
$\int_0^2{\sqrt{x+\sqrt{x+\sqrt{x+...}}}}}dx$

Let $X=\sqrt{x+\sqrt{x+\sqrt{x+\cdots}}}, x\geq 0$ . Note $X\geq0$ .

$X^2 = x + \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} \Rightarrow X^2 = x+X$ .

So $X^2 - X -x=0\Rightarrow X = \frac{1\pm \sqrt{1+4x}}{2}$ .

As $X\geq 0$ , $X = \frac{1+ \sqrt{1+4x}}{2}$ .

So $I = \int _0 ^2 \sqrt{x+\sqrt{x+\sqrt{x+\cdots}}} \text{ d}x = \int _0 ^2 \frac{1+ \sqrt{1+4x}}{2} \text{ d}x$

$\Rightarrow I = \frac{1}{2}\left[x + \frac{1}{4}\cdot \frac{2}{3}(1+4x)^{\frac{3}{2}} \right] _0 ^2$

$=\frac{1}{2}\left(\left(2 + \frac{1}{6}(1+8)^\frac{3}{2} \right) -\left(0 + \frac{1}{6}\cdot 1^{\frac{3}{2}} \right) \right)$

$=\frac{1}{2}\left(\left(2 +\frac{27}{6} \right) - \frac{1}{6}\right)$

$=\frac{19}{6}$ .

InteGrand · Apr 7, 2015

Re: MX2 2015 Integration Marathon

$\underline{NEW QUESTION}$

$$Find $I= \int \frac{169 \sin x}{5\sin x + 12\cos x}\text{ d}x$.$

simpleetal · Apr 7, 2015

Re: MX2 2015 Integration Marathon

InteGrand said:
$\underline{NEW QUESTION}$

$$Find $I= \int \frac{169 \sin x}{5\sin x + 12\cos x}\text{ d}x$.$

After a few manipulations with t form, the expression becomes 338t/((1+t^2)((5t+12)) dt(unless I made a silly somewhere...but u get the idea) which u can then integrate by using partial fractions.

Ekman · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Next Question:

$\int \frac{x}{\sqrt{(x^2+x)^3}} dx$

SpiralFlex · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
After a few manipulations with t form, the expression becomes 338t/((1+t^2)((5t+12)) dt(unless I made a silly somewhere...but u get the idea) which u can then integrate by using partial fractions.

Usually I don't think someone would post a question with just a straight up 4U technique, it usually requires some extra work/substitution.

Notice how 169 = 12^2 + 5^2

simpleetal · Apr 7, 2015

Re: MX2 2015 Integration Marathon

SpiralFlex said:
Usually I don't think someone would post a question with just a straight up 4U technique, it usually requires some extra work/substitution.

Notice how 169 = 12^2 + 5^2

Actually, contrary to what you think, my method does work. Of course what you stated is probably a great alternative, but ur post sounds as if my method was invaild, when it clearly works

SpiralFlex · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
Actually, contrary to what you think, my method does work. Of course what you stated is probably a great alternative, but ur post sounds as if my method was invaild, when it clearly works

Your method is valid. I'm not attacking your method, it's fine. I'm saying there would be an alternative since in this thread people usually post things you can't do directly, i.e using integration by parts straight away.

simpleetal · Apr 7, 2015

Re: MX2 2015 Integration Marathon

SpiralFlex said:
Your method is valid. I'm not attacking your method, it's fine. I'm saying there would be an alternative since in this thread people usually post things you can't do directly, i.e using integration by parts straight away.

Yeah sorry I might've sounded rude/aggressive cause you know it's the internet but I didn't mean to be rude lol. But yeah I see ur point sir

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
Yeah sorry I might've sounded rude/aggressive cause you know it's the internet but I didn't mean to be rude lol. But yeah I see ur point sir

shes a girl

SpiralFlex · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
Yeah sorry I might've sounded rude/aggressive cause you know it's the internet but I didn't mean to be rude lol. But yeah I see ur point sir

Np.

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
oh lol didn't see this question lol.

Use the subsitution u= x^2 + x to convert expression into du/(2u^(3/2)) , which is simple to integrate

lol notsure about your solution i think im going to post mine

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Next Question:

$\int \frac{x}{\sqrt{(x^2+x)^3}} dx$

$$ Construct triangle with lengths: $ 1, \sqrt{x} , \sqrt{1+x}$

$sin\theta= \sqrt{x}$

$dx=2sin\theta cos\theta d\theta$

$cos\theta= \frac{1}{\sqrt{1+x}}$

$\therefore $ question becomes: $ \int{2cos^4\theta}d\theta$
EDIT : my bad sin\theta isnt rootx but tan theta is
$rest is pretty straight foward$

$\int{\frac{dx}{\sqrt{x} (x+1)^{3/2}}}$

$tan\theta= \sqrt{x}$

$2tan\theta sec^2\theta d\theta=dx$

$cos\theta = \frac{1}{\sqrt{1+x}}$

$= \int{2cos\theta}d\theta$

$= \frac{2\sqrt{x}}{\sqrt{1+x}} +C$

Sy123 · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$$ Construct triangle with lengths: $ 1, \sqrt{x} , \sqrt{1+x}$

$sin\theta= \sqrt{x}$

$dx=2sin\theta cos\theta d\theta$

$cos\theta= \frac{1}{\sqrt{1+x}}$

$\therefore $ question becomes: $ \int{2cos^4\theta}d\theta$
EDIT : my bad sin\theta isnt rootx but tan theta is
$rest is pretty straight foward$

Since you will want x > 0, then sqrt(1+x) > 1, meaning it is the hypotenuse, and so sin theta =/= sqrt(x) (assuming theta is an angle in the triangle which is right angled)

braintic · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$$ Construct triangle with lengths: $ 1, \sqrt{x} , \sqrt{1+x}$

$sin\theta= \sqrt{x}$

$dx=2sin\theta cos\theta d\theta$

$cos\theta= \frac{1}{\sqrt{1+x}}$

$\therefore $ question becomes: $ \int{2cos^4\theta}d\theta$
EDIT : my bad sin\theta isnt rootx but tan theta is
$rest is pretty straight foward$

Wouldn't tanθ be √x [since √(1+x) is clearly the hypotenuse]

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

$\int{\frac{dx}{\sqrt{x} (x+1)^{3/2}}}$

$tan\theta= \sqrt{x}$

$2tan\theta sec^2\theta d\theta=dx$

$cos\theta = \frac{1}{\sqrt{1+x}}$

$= \int{2cos\theta}d\theta$
$= \frac{2\sqrt{x}}{\sqrt{1+x}} +C$

simpleetal · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Lol why are people only active on this marathon? Can someone plz post a problem in the other one (advanced)?

Drsoccerball · Apr 7, 2015

Re: MX2 2015 Integration Marathon

simpleetal said:
Lol why are people only active on this marathon? Can someone plz post a problem in the other one (advanced)?

2 unit or advanced 4u ?

simpleetal · Apr 7, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
2 unit or advanced 4u ?

advanced 4u I think. They have some very interesting problems

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