2008 HSC Reduction Formulae Q. (1 Viewer)

K4M1N3 · Jun 3, 2011

For $n\geqslant 0$ , Let:

$I_{n}=\int_{0}^{\pi/4}tan^2^n\theta \; d\theta$

For $n\geqslant 1$ Show that:

$I_n=\frac{1}{2n-1}-I_n_-_1$

Could someone please show me how to show this...been trying for a while but can't seem to squeeze it out.

Drongoski · Jun 3, 2011

$I_n = \int ^{\frac {\pi}{4}} _0 tan^{2(n-1)}\theta (sec^2 \theta - 1) d \theta$

$= \int _0 ^ {\frac {\pi}{4}} tan ^{2(n-1)} \theta dtan \theta - \int _0 ^{}\frac {\pi}{4} tan ^ {2(n-1)} \theta d \theta$

$= \frac {tan ^{2n-1}}{2n-1} ]^{\frac {\pi}{4}} _0 - I_{n-1}$

$= \frac {1}{2n-1} - I_{n-1}$

subject to n > 0

K4M1N3 · Jun 3, 2011

(-_-) so simple.....thanks.

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2008 HSC Reduction Formulae Q. (1 Viewer)

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