Integration question (1 Viewer)

hermine · Aug 15, 2009

$I_{n}=\int_{0}^{\Pi /4} tan^{2n}x\: dx, n\geq 1, show\, I_{n}+I_{n-2}=\frac{1}{2n-1}$

I made tan^2n(x) = [tan^2(x)]^n then = [tan^2(x)]^n-1 . tan^2(x)
but then got stuck lol, thaanks in advance

sidereal · Aug 15, 2009

are you sure it's not $I_{n-1}$ ? i checked it with mathematica and n-2 doesn't work, n-1 does.

$I_n = \int_0^{\pi/4} \tan^{2n} x\,\text{d}x\\ = \int_0^{\pi/4} \tan^{2n-2}x (\sec^2 x - 1)\,\text{d}x \\ = \int_0^{\pi/4} \tan^{2n-2}x .\sec^2 x\,\text{d}x - \int_0^{\pi/4} \tan^{2(n-1)}x\,\text{d}x \\ = \left[\frac{1}{2n-1}\tan^{2n-1}x \right ]_0^{\pi/4} - I_{n-1} \\ = \frac{1}{2n-1} - I_{n-1} \text{ where } n\ge1 \\ \therefore I_n + I_{n-1} = \frac{1}{2n-1}$

hermine · Aug 15, 2009

sidereal said:
are you sure it's not $I_{n-1}$ ?

hmm, its definately $I_{n-2}$ , from the Sydney boys high trial paper 2004 :S

sidereal · Aug 15, 2009

looks like a 1 to me >_>

Revacious · Aug 15, 2009

i can confirm it is a 1.

Aerath · Aug 15, 2009

I've got an original and it's n-1....

hermine · Aug 15, 2009

My copy has a 2 what the effff bro, anyways thanks for that lol

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Integration question (1 Viewer)

hermine

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