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

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

braintic said:
Umm ...... log 0 and log 1 are a bit of a problem ....

It works on wolfram giving a definite answer

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Is there a method of doing it such that it is still within the bounds of the MX2 syllabus?

Ekman · Apr 3, 2015

Re: MX2 2015 Integration Marathon

tomatoplant said:
$\int_{0}^{6} $ (6-x) $ \sqrt {12x - x^{2}}$

$$Let u=$12x-x^2 \therefore du=(12-2x)dx$

$$Hence: $ \int_{0}^{36} \frac{\sqrt{u}}{2} = [\frac{\sqrt{u^3}}{3}]_{0}^{36} =72$

studynerd said:
Ahahahahaha [emoji23] [emoji23] [emoji23]

But.. ignoring the boundaries.. is it possible to find the primitive with only mx2 knowledge? I'm stuck :/

So am I, I feel like it uses uni maths....

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
$$Let u=$12x-x^2 \therefore du=(12-2x)dx$

$$Hence: $ \int_{0}^{36} \frac{\sqrt{u}}{2} = [\frac{\sqrt{u^3}}{3}]_{0}^{36} =72$

So am I, I feel like it uses uni maths....

Its a university question yes

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
Is there a method of doing it such that it is still within the bounds of the MX2 syllabus?

ill give it a go

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

NEW QUESTION:

$Evaluate $\int \sqrt{1+\sin(x)}\cot(x)dx$$

Squar3root · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int \sqrt{1+\sin(x)}\cot(x)dx$$

let u = sin(x)

then we get integral[ sqrt(1 + u) ]

= (2/3)*(sin(x) +1)^(3/2)

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Squar3root said:
let u = sin(x)

then we get integral[ sqrt(1 + u) ]

= (2/3)*(sin(x) +1)^(3/2)

You made a slight mistake, it is says cot(x), not cos(x) so u = sin(x) won't necessarily work as a valid substitution.

Squar3root · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
You made a slight mistake, it is says cot(x), not cos(x) so u = sin(x) won't necessarily work as a valid substitution.

oops, my bad

I got log(sin(x)) + 2*sqrt(sin(x))

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
NEW QUESTION:

$Evaluate $\int \sqrt{1+\sin(x)}\cot(x)dx$$

$$ let $ u=1+sinx$
$du=cosx dx$

$\int{\frac{\sqrt{u}}{u-1}}du$

$$ Triangle with sides $ : 1,u^{1/2}, \sqrt{-(u-1)}$

$\frac{-1}{cos^2\theta} = u-1$

$tan\theta = \sqrt{u}$

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

$\therefore = \int{-2(sec^2\theta -1)}d\theta$

$-2 tan\theta +2\theta + C$

$-2 \sqrt{1+ sinx} +2tan^{-1} (\sqrt{1+sinx}}) + C$

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Squar3root said:
oops, my bad

I got log(sin(x)) + 2*sqrt(sin(x))

Still ain't right

Squar3root · Apr 3, 2015

Re: MX2 2015 Integration Marathon

working:

Squar3root · Apr 3, 2015

Re: MX2 2015 Integration Marathon

I don't see what I'm doing wrong

Ekman · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Squar3root said:
I don't see what I'm doing wrong

Invalid split of partial fractions. You can test it out by converging the two fractions together, and you wont get the original equation.

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

How did you get from 2nd to 3rd line?

Squar3root · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Ekman said:
Invalid split of partial fractions. You can test it out by converging the two fractions together, and you wont get the original equation.

oh

I see

VBN2470 said:
How did you get from 2nd to 3rd line?

bad math

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
How did you get from 2nd to 3rd line?

is my solution correct?

VBN2470 · Apr 3, 2015

Re: MX2 2015 Integration Marathon

No. Try again.

studynerd · Apr 3, 2015

Re: MX2 2015 Integration Marathon

Drsoccerball said:
$$ let $ u=1+sinx$
$du=cosx dx$

$\int{\frac{\sqrt{u}}{u-1}}du$

$$ Triangle with sides $ : 1,u^{1/2}, \sqrt{-(u-1)}$

$\frac{-1}{cos^2\theta} = u-1$

$tan\theta = \sqrt{u}$

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

$\therefore = \int{-2(sec^2\theta -1)}d\theta$

$-2 tan\theta +2\theta + C$

$-2 \sqrt{1+ sinx} +2tan^{-1} (\sqrt{1+sinx}}) + C$

I think this way is faster.. or maybe i just dont like drawing triangles..

let u^2=1+sin^2.... make cot(x)=cos(x)/sin(x)... the you can simply put in the 2udu... it all falls out after that

Drsoccerball · Apr 3, 2015

Re: MX2 2015 Integration Marathon

VBN2470 said:
No. Try again.

Can you see my mistake

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