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HSC Stuff => HSC Maths Stuff => HSC Subjects + Help => HSC Mathematics Extension 2 => Topic started by: stanleypeng on July 14, 2017, 06:44:32 pm

Title: Integration Question
Post by: stanleypeng on July 14, 2017, 06:44:32 pm: Hey, this is my first time posting a question so apologies if I'm doing this wrong. Could you please help with part ii), I did part i) and started to use the formula but got stuck halfway. Thanks :)

Mod edit: Nothing wrong with posting the question, however if you prefer to not post in the question thread please make your thread title clearer.
Title: Re: Integration Question
Post by: RuiAce on July 14, 2017, 07:48:56 pm: $\begin{align*}\int_0^{\frac{\pi}{4}}\frac{1-\tan x}{1+\tan x}dx&=\int_0^{\frac{\pi}{4}}\frac{1-\tan \left(\frac{\pi}{4}-x\right)}{1+\tan \left(\frac{\pi}{4}-x\right)}\\ &= \int_0^{\frac{\pi}{4}}\frac{1-\frac{1-\tan x}{1+\tan x}}{1+\frac{1-\tan x}{1+\tan x}}\tag{compound angle}\\ &= \int_0^{\frac{\pi}{4}}\frac{(1+\tan x)-(1-\tan x)}{(1+\tan x)+(1-\tan x)}\\ &= \int_0^{\frac{\pi}{4}}\tan x\, dx \end{align*}\\ \text{The rest I leave as your exercise}\\ \text{because you should be well familiar as to how to integrate tan}$
Title: Re: Integration Question
Post by: jamonwindeyer on July 14, 2017, 10:14:28 pm: Quote from: stanleypeng on July 14, 2017, 06:44:32 pm
Hey, this is my first time posting a question so apologies if I'm doing this wrong. Could you please help with part ii), I did part i) and started to use the formula but got stuck halfway. Thanks :)

Mod edit: Nothing wrong with posting the question, however if you prefer to not post in the question thread please make your thread title clearer.

Welcome to the forums Stanley! Be sure to let us know if we can help you get your bearings ;D
Title: Re: Integration Question
Post by: stanleypeng on July 15, 2017, 12:33:15 am: I have another question about integration but I don't know how to post it under a thread so I'm posting it as a reply instead. Again, I completed part i) but need help with part ii), thanks :).
Title: Re: Integration Question
Post by: RuiAce on July 15, 2017, 08:46:03 am: Quote from: stanleypeng on July 15, 2017, 12:33:15 am
I have another question about integration but I don't know how to post it under a thread so I'm posting it as a reply instead. Again, I completed part i) but need help with part ii), thanks :).
$\text{For example, }I_{n-2}=\frac{n-3}{n-2}I_{n-4}$
$\text{So by repeatedly applying the reduction formula}\\ \begin{align*}I_{2n}&=\frac{2n-1}{2n}I_{2n-2}\\ &= \frac{(2n-1)(2n-3)}{2n(2n-2)}I_{2n-4}\\ &= \frac{(2n-1)(2n-3)(2n-5)}{2n(2n-2)(2n-4)}I_{n-6}\\ &\vdots \\ &= \frac{(2n-1)(2n-3)\dots(3)(1) }{2n(2n-2)\dots (4)(2)}I_0\end{align*}$
$\text{The numerator needs to be }(2n)!,\text{ but that's }(2n)(2n-1)\dots (3)(2)(1)\\ \text{So we multiply the top and bottom to force its way in.}\\ \begin{align*}I_{2n}&=\frac{2n(2n-1)(2n-2)(2n-3)\dots(4)(3)(2)(1)}{(2n)^2(2n-2)^2\dots (4)^2(2)^2}I_0\\ &= \frac{(2n)!}{(2n(2n-2)\dots (4)(2))^2}I_0\end{align*}$
$\text{The denominator needs to be }(n!)^2\\ \text{but right now our terms are all double in size because we need }n(n-1)\dots (2)(1)\\ \text{It can be hard to see that }(2n)(2n-2)\dots(4)(2)=2^{n-1} (n)(n-1)\dots(2)(1)\\ \text{So we now have }I_n=\frac{(2n)!}{2^{n-1} (n!)^2}I_0$
Carefully note that there are n-1 terms in (2n)(2n-2)...(2), not n terms.
$\text{But by direct computation }I_0=\frac{\pi}{4}\text{ so the last bit is easy}$