HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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Very nice integral

dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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Rushed this one, hopefully no errors

dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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easier than I thought

dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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dunjaaa · Mar 21, 2014

Re: MX2 Integration Marathon

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mind the misuse of integral

Sy123 · Mar 21, 2014

Re: MX2 Integration Marathon

All of them seems good to me

Sy123 · Mar 21, 2014

Re: MX2 Integration Marathon

$\int \sin (\ln x) + \cos (\ln x) \ dx$

dunjaaa · Mar 22, 2014

Re: MX2 Integration Marathon

xsin(ln(x))+C reverse product rule

Sy123 · Mar 22, 2014

Re: MX2 Integration Marathon

Hence

$\int \sin (\ln x) \ dx$

dunjaaa · Mar 23, 2014

Re: MX2 Integration Marathon

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Sy123 · Mar 23, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30172

I guess you can do that, though when I said Hence, I was thinking of:

$\int \sin(\ln(x)) + \cos( \ln x) \ dx = x\sin \ln x + c$

$\int \sin \ln x + \cos \ln x \ dx \ = \sqrt{2} \int \sin(\ln x + \frac{\pi}{4}) \ dx \\ u = e^{\pi/4} x \\ \\ = \frac{\sqrt{2}}{e^{\pi/4}} \int \sin \ln u \ du = x\sin \ln x + c$

$\therefore \ \int \sin \ln u \ du = \frac{e^{\pi/4}}{\sqrt{2}} \cdot \frac{u}{e^{\pi/4}} \sin (\ln u - \frac{\pi}{4}) = \sqrt{2} u \sin \left(\ln u - \frac{\pi}{4}\right)$

Note that in an exam however you will not be marked if you do not use the previous part

dunjaaa · Mar 23, 2014

Re: MX2 Integration Marathon

Oh, well the first thing that sprung in my head was IBP twice lol

Sy123 · Mar 24, 2014

Re: MX2 Integration Marathon

$\int_{-1}^1 \tan^{-1}(e^x) \ dx$

Sy123 · Mar 24, 2014

Re: MX2 Integration Marathon

$\int_{0}^{\pi/2} \frac{dx }{a^2\sin^2x + b^2\cos^2x}$

Carrotsticks · Mar 24, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_{-1}^1 \tan^{-1}(e^x) \ dx$

Haha I like this one, assuming you intended the method I have in mind.

RealiseNothing · Mar 24, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_{-1}^1 \tan^{-1}(e^x) \ dx$

$f(x) = \tan^{-1}(e^x)$

$f(-x) = \tan^{-1}(\frac{1}{e^x})$

$f(x)+f(-x)=\frac{\pi}{2}$ (just consider a right angled triangle)

So we can just make the curve $=\frac{\pi}{2}$ and change the limits from 0 to 1 and the answer is $\frac{\pi}{2}$

What I'm doing is "cutting" the section of the graph between -1 and 0, and "pasting" it on top of the graph between 0 and 1.

Sy123 · Mar 25, 2014

Re: MX2 Integration Marathon

RealiseNothing said:
$f(x) = \tan^{-1}(e^x)$

$f(-x) = \tan^{-1}(\frac{1}{e^x})$

$f(x)+f(-x)=\frac{\pi}{2}$ (just consider a right angled triangle)

So we can just make the curve $=\frac{\pi}{2}$ and change the limits from 0 to 1 and the answer is $\frac{\pi}{2}$

What I'm doing is "cutting" the section of the graph between -1 and 0, and "pasting" it on top of the graph between 0 and 1.

Yep, the more technical approach though would be to just set u=-x and apply the same formula.

Sy123 said:
$\int_{0}^{\pi/2} \frac{dx }{a^2\sin^2x + b^2\cos^2x}$

Bump

don't use t-sub plz

RealiseNothing · Mar 25, 2014

Re: MX2 Integration Marathon

Sy123 said:
Yep, the more technical approach though would be to just set u=-x and apply the same formula.

don't use t-sub plz

HSC 2012-14 MX2 Integration Marathon (archive) (1 Viewer)

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