Recursion formula for integrals of tan^n(θ)

[BLUE_CHIP]

Someone fix my tex pls...


I'm only 16 so i put this on the K-12 forum but they can't seem to help...

OK. I've had a little break from my studdies and need some help with this...

[tex]I_n(x)=\int\limits_0^x \tan^n{{\theta}}{{d\theta}},n\leq{0},{{x}}<\frac{\pi}{2}[\tex]<br /> <br /> By writing [tex]\tan{\theta}[\tex] as [tex]\tan^{n-2}{\theta}\tan^2{\theta}[\tex], or otherwise, show that<br /> <br /> [tex]I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x<\frac{\pi}{2}[\tex]<br /> <br /> Hence evaluate [tex]\int\limits_{0}^{\frac{\pi}{3}}\tan^4{\thet<br /> a}d\theta[\tex], leaving your answers in terms of $$\pi[\tex]<br /> <br /> Thanks (Goddam further maths)$$[/tex][/tex][/tex][/tex][/tex]

 

[lethe]

Originally posted by BLUE_CHIP
Someone fix my tex pls...


I'm only 16 so i put this on the K-12 forum but they can't seem to help...

OK. I've had a little break from my studdies and need some help with this...

$$I_n(x)=\int\limits_0^x \tan^n{{\theta}}{{d\theta}},n\leq{0},{{x}}<\frac{\pi}{2}$$

By writing $$\tan{\theta}$$ as $$\tan^{n-2}{\theta}\tan^2{\theta}$$, or otherwise, show that

$$I_n(x)=\frac{1}{n-1}\tan^{n-1}{x}-I_{n-2}(x), n\leq{2},x<\frac{\pi}{2}$$

Hence evaluate $$\int\limits_{0}^{\frac{\pi}{3}}\tan^4{\thet<br /> a}d\theta$$, leaving your answers in terms of $$\pi$$

Thanks (Goddam further maths)

 

[lethe]

bulletin board commands like [ /tex ], get slashes. LaTeX commands get backslashes like \Sum

OK. here we go:

$$\begin{gather*}<br /> \tan^n \theta=\tan^{n-2}\theta \tan^2 \theta=\\<br /> \tan^{n-2}\theta(\sec^2\theta-1)=\\<br /> \tan^{n-2}\theta\sec^2\theta-\tan^{n-2}\theta<br /> \end{gather*}$$

so

$$\int\tan^2\theta\ d\theta=\int\tan^{n-2}\theta\sec^2\theta\ d\theta-\int\tan^{n-2}\theta\ d\theta$$
use a [itex]u=\tan\theta[/tex] substitution and you have<br /> $$I_n(x)=\int^{u(x)} u^{n-2}\ du-I_{n-2}(x)$$<br /> <br /> and maybe you can take it from there?[/itex]

 

[joc]

BLUE_CHIP: you're taking Further Maths at 16? as in the A-level subject? that's pretty impressive...

 

[BLUE_CHIP]

Well, I'm taking the A-level this year but I had done all the single Maths before so my teacher said that we should start on P4 and P5 so HeyHo. Fun and games...

 

[joc]

heh that's cool. then you'll be like, a match for some of the more accelerated people in the US :P enjoy yourself.

 

[PrudensOptimus]

Lame, just another "Hey look I'm 16 and I'm integrating, but I don't know what to do, pls help and btw, I'm 16, say I'm cool pls" thread...

STFU pls. thanks.

 

[franznietzsche]

Originally posted by PrudensOptimus
Lame, just another "Hey look I'm 16 and I'm integrating, but I don't know what to do, pls help and btw, I'm 16, say I'm cool pls" thread...

STFU pls. thanks.

dude what is your problem? get off the guy's case. Seriously, so what if he mentions he's sixteen and integrating, for most people (excepting the true geniuses) that's something of an accomplishment. so lay off with being such an ass to someone just looking for help.

 

[joc]

actually integration in itself isn't particularly impressive (most people around me learn it at 16); in fact i just recalled that taking Further Math at 16-17 is actually normal and not exceptional, so i take my compliment back.

no offence blue_chip :)

 

[franznietzsche]

well i live in california, and here about 80 (out of 3000) students a year take AP calculus, while about 70% can't pass a test on simple algebra and geometry. So for this educationally challenged state it is something of an accomplishment. And even if it isn't its still no reason to go off on him.

 

[himanshu121]

Well guys Here in India we have these kind of functions and problems when we are 17
and calculus is dominating feature. I must say it is compulsory here.