# Thread: HSC 2017 MX2 Integration Marathon

1. ## Re: HSC 4U Integration Marathon 2017

$\noindent Evaluate \int_{0}^{\frac{\pi}{2}}\frac{1}{3+5\cos x} dx$

2. ## Re: HSC 4U Integration Marathon 2017

you can rearrange gif.latex.gifand gif.latex.gif to simplify the inside the integrals to get

gif.latex.gif
use substitution gif.latex.gif
then gif.latex.gif

gif.latex.gif

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let gif.latex.gif

gif.latex.gif constant

From the you can see the integrals is tan inverse of \sqrt{u^2-1}
We can back track to find it in terms of theta

3. ## Re: HSC 4U Integration Marathon 2017

Learn Latex pls

4. ## Re: HSC 4U Integration Marathon 2017

$\int{\frac{x^2 -1}{x^2 + 1} \cdot \frac{1}{\sqrt{1+x^4}}}dx$

5. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Drsoccerball
$\int{\frac{x^2 -1}{x^2 + 1} \cdot \frac{1}{\sqrt{1+x^4}}}dx$
smh...

6. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by Drsoccerball
$\int{\frac{x^2 -1}{x^2 + 1} \cdot \frac{1}{\sqrt{1+x^4}}}dx$
$\noindent I= \int \frac{x^2-1 }{(x^2+1)\sqrt{1+x^4}} dx=\int \frac{x^2-1}{x(x^2+1)\sqrt{x^2+\frac{1}{x^2}}} dx \\ dividing the denominator and the numerator by x we get \\\\ I =\int \frac{x-\frac{1}{x}}{x(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^ 2}}} dx=\int \frac{\left ( \frac{x^2-1}{x^2} \right )}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}} dx \\ let u=x+\frac{1}{x} \Rightarrow du=1-\frac{1}{x^2} dx \\ \\\therefore dx=\frac{x^2}{x^2-1} du. \\\\ Substituing in the dx, the numerator cancels out.\\ I=\int \frac{1}{u\sqrt{u^2-2}} du \\\\ let v=\sqrt{u^2-2}\Rightarrow du = \frac{\sqrt{u^2-2}}{u} dv \\ subbing in, the integral becomes \int \frac{1}{v^2+2} dv= \frac{1}{\sqrt{2}}\tan^{-1}\frac{v}{\sqrt{2}}+C$

$\noindent Reversing all the substitutions, we'll get \\ I=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{u^2-2}}{\sqrt{2}}+C = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}+C\\\\\ \\\ simplifying further, it'll become \frac{1}{\sqrt{2}}\tan^{-1}\left ( \frac{\sqrt{1+x^4}}{x\sqrt{2}} \right )+C$

7. ## Re: HSC 4U Integration Marathon 2017

$\int_{0}^{1}\frac{1}{(x+3)^2}\ln \left(\frac{x+2}{x+1}\right)\;\textrm{d}x$

8. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by jathu123
$\noindent I= \int \frac{x^2-1 }{(x^2+1)\sqrt{1+x^4}} dx=\int \frac{x^2-1}{x(x^2+1)\sqrt{x^2+\frac{1}{x^2}}} dx \\ dividing the denominator and the numerator by x we get \\\\ I =\int \frac{x-\frac{1}{x}}{x(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^ 2}}} dx=\int \frac{\left ( \frac{x^2-1}{x^2} \right )}{(x+\frac{1}{x})\sqrt{x^2+\frac{1}{x^2}}} dx \\ let u=x+\frac{1}{x} \Rightarrow du=1-\frac{1}{x^2} dx \\ \\\therefore dx=\frac{x^2}{x^2-1} du. \\\\ Substituing in the dx, the numerator cancels out.\\ I=\int \frac{1}{u\sqrt{u^2-2}} du \\\\ let v=\sqrt{u^2-2}\Rightarrow du = \frac{\sqrt{u^2-2}}{u} dv \\ subbing in, the integral becomes \int \frac{1}{v^2+2} dv= \frac{1}{\sqrt{2}}\tan^{-1}\frac{v}{\sqrt{2}}+C$

$\noindent Reversing all the substitutions, we'll get \\ I=\frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{u^2-2}}{\sqrt{2}}+C = \frac{1}{\sqrt{2}}\tan^{-1}\frac{\sqrt{x^2+\frac{1}{x^2}}}{\sqrt{2}}+C\\\\\ \\\ simplifying further, it'll become \frac{1}{\sqrt{2}}\tan^{-1}\left ( \frac{\sqrt{1+x^4}}{x\sqrt{2}} \right )+C$
Somewhat convoluted.

$\noindent \int \frac{x^2-1}{x^2+1} \frac{\text{d}x}{\sqrt{1+x^4}} = \int \frac{\left(1-\frac{1}{x^2}\right) \text{d}x}{\left( x+\frac{1}{x} \right) \sqrt{x^2+\frac{1}{x^2}}} \\\\ = \int \frac{\text{d} \left( x+\frac{1}{x} \right)}{\left( x+\frac{1}{x} \right) \sqrt{\left( x+\frac{1}{x} \right)^2 -2}} = \frac{1}{\sqrt{2}} \sec^{-1}{\left( \frac{x+\frac{1}{x}}{\sqrt{2}} \right)} + \mathcal{C}$

9. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by leehuan
smh...
Why that reaction haha? Is there some story behind this integral?

10. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
Why that reaction haha? Is there some story behind this integral?
Seen it before a bit too many times :P

11. ## Re: HSC 4U Integration Marathon 2017

$\noindent Find \int _{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x +\sin x}\, \mathrm{d}x.$

12. ## Re: HSC 4U Integration Marathon 2017

$\noindent Let \lambda , t > 0. Show that \int_{0}^{t} \frac{\lambda^{n}}{(n-1)!}\tau^{n-1} e^{-\lambda \tau}\, \mathrm{d}\tau = 1 - \sum_{k=0}^{n-1} e^{-\lambda t}\frac{\left(\lambda t\right)^{k}}{k!}, for n=1,2,3,\ldots.$

13. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Find \int _{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x +\sin x}\, \mathrm{d}x.$
$\noindent let u=\frac{\pi}{2}-x \Rightarrow dx = -du. When x=\frac{\pi}{2}, u=0. When x=0, u=\frac{\pi}{2}.\\\therefore I= \int_{0}^{\frac{\pi}{2}}\frac{\cos x}{\cos x+\sin x} dx=-\int_{\frac{\pi}{2}}^{0}\frac{\cos (\frac{\pi}{2}-u)}{\cos (\frac{\pi}{2}-u)+\sin (\frac{\pi}{2}-u) } du \\ = \int_{0}^{\frac{\pi}{2}}\frac{\sin u}{\sin u+\cos u} du=\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x} dx\\\\ \therefore 2I=\int_{0}^{\frac{\pi}{2}}\frac{ \sin x+\cos x}{\sin x + \cos x} dx =\frac{\pi}{2}\\ \\ \therefore I=\frac{\pi}{4}$

14. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Let \lambda , t > 0. Show that \int_{0}^{t} \frac{\lambda^{n}}{(n-1)!}\tau^{n-1} e^{-\lambda \tau}\, \mathrm{d}\tau = 1 - \sum_{k=0}^{n-1} e^{-\lambda t}\frac{\left(\lambda t\right)^{k}}{k!}, for n=1,2,3,\ldots.$
don't scare off the children

15. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by abecina
$\int_{0}^{1}\frac{1}{(x+3)^2}\ln \left(\frac{x+2}{x+1}\right)\;\textrm{d}x$
http://imgur.com/a/LQn6t

Checked it on wolfram so it's all good.

16. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Find \int _{0}^{\frac{\pi}{2}} \frac{\cos x}{\cos x +\sin x}\, \mathrm{d}x.$
Also this is another method: http://imgur.com/h2IBmvL

17. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by calamebe
http://imgur.com/a/LQn6t

Checked it on wolfram so it's all good.
that is not in it's simplest form.

18. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Let \lambda , t > 0. Show that \int_{0}^{t} \frac{\lambda^{n}}{(n-1)!}\tau^{n-1} e^{-\lambda \tau}\, \mathrm{d}\tau = 1 - \sum_{k=0}^{n-1} e^{-\lambda t}\frac{\left(\lambda t\right)^{k}}{k!}, for n=1,2,3,\ldots.$
http://imgur.com/d7oqSIR

19. ## Re: HSC 4U Integration Marathon 2017

Also sorry if my handwriting poses any issues

20. ## Re: HSC 4U Integration Marathon 2017

that is not in it's simplest form.
I know, but it's the right answer and I can't be bothered to put it in it's simplest form.

Edit: stuff it ln(3^(5/4)/2^(23/12))

21. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by InteGrand
$\noindent Let \lambda , t > 0. Show that \int_{0}^{t} \frac{\lambda^{n}}{(n-1)!}\tau^{n-1} e^{-\lambda \tau}\, \mathrm{d}\tau = 1 - \sum_{k=0}^{n-1} e^{-\lambda t}\frac{\left(\lambda t\right)^{k}}{k!}, for n=1,2,3,\ldots.$
Wait what... is there something special about the Poisson distribution I should know

22. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by leehuan
Wait what... is there something special about the Poisson distribution I should know
Leehuan this isn't the extracurricular marathon take your stats elsewhere yo

23. ## Re: HSC 4U Integration Marathon 2017

Meh doesn't work anyway, cause if it were Poisson the upper limit of the sum should be inf

24. ## Re: HSC 4U Integration Marathon 2017

Originally Posted by leehuan
Wait what... is there something special about the Poisson distribution I should know
The LHS integrates the PDF of a Gamma(n,beta) r.v. (where beta = 1/lambda) from 0 to t and the RHS is the probability a Poisson(lambda*t) r.v. is bigger than or equal to n. The identity thus shows that in a Poisson process, the number of events occurring in time 0 to t when the rate is lambda is Poisson (lambda*t) if we assume the time between events is iid exponential(lambda) (noting that a sum of n iid exponential(lambda) is Gamma(n, (1/lambda)).

25. ## Re: HSC 4U Integration Marathon 2017

don't scare off the children
Find

$\int \frac { x^{ \frac { 2 }{ 3 } } }{ \left( 1+x^{ \frac { 3 }{ 5 } } \right) ^{ 2 } }\, \mathrm{d}x$.

(Courtesy of leehuan here: http://community.boredofstudies.org/...ml#post7183192 )

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