HSC Mathematics Marathon (5 Viewers)

bored of fail 1 · Feb 16, 2011

you like multiply top and bottom in the sqrt by (16+x) , then use a substitution on the difference of squares term on the bottom

Drongoski · Feb 16, 2011

deterministic said:
You mean:
$\int \sqrt{\frac{x}{16-x}} dx$

Try using the substitution: $x = 16sin^2 \theta$

deterministic · Feb 16, 2011

^Should be easy after the above substitution.

Next question:
$\int sin(2x)ln(sinx) dx$

bored of fail 1 · Feb 16, 2011

deterministic said:
^Should be easy after the above substitution.

Next question:
$\int sin(2x)ln(sinx) dx$

= 2sin(x)cos(x) ln ( sin(x) )

let u=sin(x) du = cos(x) dx

so integral =2u ln(u) du

then use integration by parts

lol no one on this site even does 4unit, just uni students answering all the questions

Drongoski · Feb 16, 2011

deterministic said:
^Should be easy after the above substitution.

Next question:
$\int sin(2x)ln(sinx) dx$

$= - \frac{1}{2} \int ln(sinx) dcos 2x$

$= -\frac{1}{2} cos2x \cdot ln \vert sin x \vert + \frac {1}{2} \int cos 2x \cdot d ln (sin x)$

$= - \frac {1}{2} cos 2x \cdot ln \vert sin x \vert + \frac {1}{2} \int cos 2x \times \frac {cos x}{sin x} dx$

$= - \frac {1}{2} cos 2x \cdot ln \vert sin x \vert + \frac {1}{2} \int (1 - 2sin^2x) \frac {cos x}{sin x} dx$

$= - \frac {1}{2} cos 2x \cdot ln \vert sin x \vert + \frac {1}{2} \int [ \frac {cos x}{sin x} - sin 2x ] dx$

$= - \frac {1}{2} cos 2x \cdot ln \vert sin x \vert + \frac {1}{2} ln \vert sin x \vert + \frac {1}{4} cos 2x + C$

$= \frac {1}{2} (1 - cos 2x) \cdot ln \vert sin x \vert + \frac {1}{4} cos 2x + C$

Wonder if this is even correct [after all the painstaking input with TeX ].

hscishard · Feb 16, 2011

bored of fail 1 said:
lol no one on this site even does 4unit, just uni students answering all the questions

There might be if you don't answer it in 2 minutes

bored of fail 1 · Feb 16, 2011

Drongoski said:
$= - \frac{1}{2} \int ln(sinx) dcos 2x$

$= -\frac{1}{2} cos2x \cdot ln \vert sin x \vert + \frac {1}{2} \int cos 2x \cdot d ln (sin x)$

$= - \frac {1}{2} ln \vert sin x \vert + \frac {1}{2} cos 2x \times \frac {cos x}{sin x} dx$

why do u complicate everything? lol

bored of fail 1 · Feb 16, 2011

bored of fail 1 said:
= 2sin(x)cos(x) ln ( sin(x) )

let u=sin(x) du = cos(x) dx

so integral =2u ln(u) du

then use integration by parts

lol no one on this site even does 4unit, just uni students answering all the questions

continuing

m = ln(u) ------ dv= u du
dm = 1/u du ----v = (1/2)u^2

so answer = 2 [ (1/2)u^2ln(u) - (1/2) integral (1/u) (u^2) du ]

= 2 [ (1/2)u^(2)ln(u) - (1/2)(1/2)u^2 ]
= 2 [ (1/2)u^2 [ ln(u) -1/2] ]
= u^2 [ln(u) - (1/2) ] +C
'

now resub u=sin(x)

doesnt look quite correct, maybe I amde mistake soemwhere

Drongoski · Feb 16, 2011

bored of fail 1 said:
why do u complicate everything? lol

Just my "weird" way of doing things. Now you do it your straightforward way.

bored of fail 1 · Feb 16, 2011

Drongoski said:
Wonder if this is even correct.

you can check on wolframa

bored of fail 1 · Feb 16, 2011

bored of fail 1 said:
you can check on wolframa

its correct

Drongoski · Feb 16, 2011

bored of fail 1 said:
you can check on wolframa

Thanks. Don't know how to.

Drongoski · Feb 17, 2011

Drongoski said:
Just my "weird" way of doing things. Now you do it your straightforward way.

By the way I was just religiously following the formula:

$\int udv = uv - \int vdu$

UnrealAnchovies · Feb 21, 2011

If a, b, y are roots of the equation $ax^3 + bx^2 + cx + d = 0$ , form an equation with roots:

$a, b, y, \frac {1}{a}, \frac {1}{b}, \frac {1}{y}$

----------

Form a cubic equation with roots a, b, y given that:

a + b + y = 4

$a^2 + b^2 + y^2 = 20$ and

$a^3 +b^3 + y^3 = 121$

ohexploitable · Feb 21, 2011

UnrealAnchovies said:
$If a, b, y are roots of the equation ax^3 + bx^2 + cx + d = 0, form an equation with roots:$

$a, b, y, \frac {1}{a}, \frac {1}{b}, \frac {1}{y}$

Fairly stock standard question.

$\\\text{Let }y=\frac{1}{\alpha}\\\alpha=\frac1y\\\text{subs. }x=\alpha\\\frac{a}{y^3}+\frac{b}{y^2}+\frac{c}{y}+d=0\\\\\left(\frac{a}{x^3}+\frac{b}{x^2}+\frac{c}{x}+d\right)(ax^3+bx+cx+d)=0\\\text{Expand and simplify.}$

UnrealAnchovies · Feb 21, 2011

ohexploitable said:
Fairly stock standard question.

$\\\text{Let }y=\frac{1}{\alpha}\\\alpha=\frac1y\\\text{subs. }x=\alpha\\\frac{a}{y^3}+\frac{b}{y^2}+\frac{c}{y}+d=0\\\\\left(\frac{a}{x^3}+\frac{b}{x^2}+\frac{c}{x}+d\right)(ax^3+bx+cx+d)=0\\\text{Expand and simplify.}$

Mhm. It's just straight out of the Terry Lee. Where do you get your actually challenging questions from?

Riproot · Feb 21, 2011

UnrealAnchovies said:
If a, b, y are roots of the equation $ax^3 + bx^2 + cx + d = 0$ , form an equation with roots:

$a, b, y, \frac {1}{a}, \frac {1}{b}, \frac {1}{y}$

The roots equal the coefficients? Very misleading question.

jyu · Feb 21, 2011

Find the relationship between a, b, c and d such that

$\frac{ab\left ( c+d \right )+cd\left ( a+b \right )+10\left ( ab+bc+cd+da \right )}{\left ( a+b \right )\left ( c+d \right )+10\left ( a+b+c+d \right )}=\frac{ab\left ( c+d \right )+cd\left ( a+b \right )+\left ( ab+bc+cd+da \right )}{\left ( a+b \right )\left ( c+d \right )+\left ( a+b+c+d \right )}$

Drongoski · Feb 21, 2011

jyu said:
Find the relationship between a, b, c and d such that

$\frac{ab\left ( c+d \right )+cd\left ( a+b \right )+10\left ( ab+bc+cd+da \right )}{\left ( a+b \right )\left ( c+d \right )+10\left ( a+b+c+d \right )}=\frac{ab\left ( c+d \right )+cd\left ( a+b \right )+\left ( ab+bc+cd+da \right )}{\left ( a+b \right )\left ( c+d \right )+\left ( a+b+c+d \right )}$

ad = bc

jyu · Feb 22, 2011

Drongoski said:
ad = bc

You've got the right answer. Care to share your solution method?

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