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

braintic · May 21, 2014

Re: MX2 Integration Marathon

abecina said:
$\int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}$
(without using complex numbers)

Solution: View attachment Integral of ln (sinx).pdf

Edit: Oops, left out a few logs. But I can't be bothered fixing it.

Ikki · May 21, 2014

Re: MX2 Integration Marathon

abecina said:
$\int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}$
(without using complex numbers)

Bloddy crazy question, started off like this is gonna get me no where, nek minnet:

Let
$I= \int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}$

$= \int_{0}^{\pi/2} \ln (\sin{(\frac{\pi}{2}-x})) \textrm{dx}$

$= \int_{0}^{\pi/2} \ln (\cos{x}) \textrm{dx}$

So I guess we could write this...

$2I= \int_{0}^{\pi/2} \ln (\sin{x}\cos{x}) \textrm{dx}$

$= \int_{0}^{\pi/2} \ln (\sin{2x})-\ln(2) \textrm{dx}$

Now let u=2x for the sin 2x integral and evaluate the ln2 one:

$= \frac{1}{2}\int_{0}^{\pi} \ln (\sin{x}) \textrm{dx}-\frac{\pi\ln(2)}{2}$

Ok, lets consider the sinx integral by itself because the boundaries pose a problem...
Let
$J= \int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx} + \int_{\pi/2}^{\pi} \ln (\sin{x}) \textrm{dx}$

For the second integral, sub u=pi-x.
$= \int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx} + \int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}$

Thus,
$J= 2\int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}$

Back to our integral question!

$2I= I-\frac{\pi\ln(2)}{2}$

$\therefore \int_{0}^{\pi/2} \ln (\sin{x}) \textrm{dx}= -\frac{\pi\ln(2)}{2}$

*Well it seems i'm a bit late LOL oh well, was worth a shot. Thanks Braintic atleast I know i'm right haha

Sy123 · May 22, 2014

Re: MX2 Integration Marathon

$\\ \int_0^{\infty} |e^{-x} \sin x| \ dx$

Ikki · May 22, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\\ \int_0^{\infty} |e^{-x} \sin x| \ dx$

*If one sketches the function inside the absolute value it is clear that the graph from 0 to infinity is positive.

Let the integral be I.

So integrating by parts using u=e^-x and dv=sinx then further intigrating by parts again we end up with:

I=[e^-x(cosx+sinx)] {0}{infinity}-I
2I=[e^-x(cosx+sinx)] {0}{infinity}
2I=1
I=1/2

*Not sure if i was assuming there...

EDIT: Yes i was, the curve must cut the axis every kpi...

dunjaaa · May 22, 2014

Re: MX2 Integration Marathon

I don't think that is the answer, because you are essentially minusing the negative region from the positive region when integrating e^-xsin(x)

dunjaaa · May 22, 2014

Re: MX2 Integration Marathon

Screen shot 2014-05-22 at 10.32.16 PM.png

Man that is one good question

Sy123 · May 22, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30481 Man that is one good question

Yep well done, something like that, using geometric series

*coughCSSA2011cough*

Sy123 · May 22, 2014

Re: MX2 Integration Marathon

$\\ \int_0^{\pi/12} \frac{dx}{(\sin x + \cos x)^4}$

dunjaaa · May 23, 2014

Re: MX2 Integration Marathon

Oh wow, condenses a 10 marker question into 1 question, typical Sy...

dunjaaa · May 23, 2014

Re: MX2 Integration Marathon

Screen shot 2014-05-23 at 12.39.51 AM.png

Alternatively:

Screen shot 2014-05-23 at 8.17.40 PM.png

changing it into cosine would make the integration less tedious and error free

Ikki · May 23, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30483

Congrats on that other question HAHA

Can you explain how you got to the second step? to the cosec^4

braintic · May 23, 2014

Re: MX2 Integration Marathon

Ikki said:
Congrats on that other question HAHA

Can you explain how you got to the second step? to the cosec^4

That's just auxiliary angle method.

Kurosaki · May 23, 2014

Re: MX2 Integration Marathon

Ikki said:
Congrats on that other question HAHA

Can you explain how you got to the second step? to the cosec^4

Most likely auxiliary angle method?

dunjaaa · May 23, 2014

Re: MX2 Integration Marathon

Auxillary angle method

Ikki · May 23, 2014

Re: MX2 Integration Marathon

So wait... Is it auxillary angle method? JKS haha
Nice job mate

Q:Prove that the derivative of arccot x=-1/(1+x^2) by considering y=arccot x.
Hence integrate 1/(1+x^2)dx.

*Idk maybe too easy lol

General Q: How would one approach this integral:
x^3/(2x-1).dx

dunjaaa · May 23, 2014

Re: MX2 Integration Marathon

Screen shot 2014-05-23 at 10.46.04 PM.png

, alternatively for (b) use polynomial division since deg(numerator)>deg(denominator)

Ikki · May 25, 2014

Re: MX2 Integration Marathon

dunjaaa said:
View attachment 30489, alternatively for (b) use polynomial division since deg(numerator)>deg(denominator)

Very nice

Thanks a lot!

Sy123 · May 26, 2014

Re: MX2 Integration Marathon

$\int_0^{\pi/8} \frac{(\cos x + \sin x)^{\frac{3}{2}} - (\cos x - \sin x)^{\frac{3}{2}}}{\sqrt{\cos 2x}} \ dx$

RealiseNothing · May 26, 2014

Re: MX2 Integration Marathon

Sy123 said:
$\int_0^{\pi/8} \frac{(\cos x + \sin x)^{\frac{3}{2}} - (\cos x - \sin x)^{\frac{3}{2}}}{\sqrt{\cos 2x}} \ dx$

$\int \frac{cosx+sinx}{\sqrt{cosx-sinx}}-\frac{cosx-sinx}{\sqrt{cosx+sinx}}$

$-2\sqrt{cosx-sinx}+2\sqrt{cosx+sinx}$

Insert limits etc.

Sy123 · May 27, 2014

Re: MX2 Integration Marathon

$\\ \int_0^{\pi /4} \frac{\sec x}{2 + \sin^2 x} \ dx$

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