O primitiva

Virgil Nicula
Mesaje: 244
Membru din: Sâm Oct 30, 2010 3:55 pm
Localitate: Bradenton, Florida

O primitiva

Mesaj de Virgil Nicula »

Sa se determine $\int\tan x\tan2x\tan 3x\ \mathrm{dx}$ .
stuart clark
Mesaje: 52
Membru din: Mie Mar 02, 2011 5:20 pm

Re: O primitiva

Mesaj de stuart clark »

Using Standard Trig. Identities

$\displaystyle \tan(3x) = \tan\left(2x+x\right) = \frac{\tan(2x)+\tan(x)}{1-\tan(2x).\tan(x)}$

$\displaystyle \tan(3x).\tan(2x).\tan(x) = \tan (3x)-\tan(2x)-\tan(x)$

So $\displaystyle \tan(3x).\tan(2x).\tan (x)dx =$$\int \tan(3x)dx-\int\tan(2x)dx-\int \tan(x)dx$

$= \frac{1}{3}\ln \mid \sec (x)\mid -\frac{1}{2}\ln \mid \sec (2xx)\mid-\ln \mid \sec(x) \mid+\mathbb{C}$
stuart clark
Mesaje: 52
Membru din: Mie Mar 02, 2011 5:20 pm

Re: O primitiva

Mesaj de stuart clark »

Using Standard Trig. Identities

$\displaystyle \tan(3x) = \tan\left(2x+x\right) = \frac{\tan(2x)+\tan(x)}{1-\tan(2x).\tan(x)}$

$\displaystyle \tan(3x).\tan(2x).\tan(x) = \tan (3x)-\tan(2x)-\tan(x)$

So $\displaystyle \int \tan(3x).\tan(2x).\tan (x)dx$

$\displaystyle = \int \tan(3x)dx-\int\tan(2x)dx-\int \tan(x)dx$

$= \frac{1}{3}\ln \mid \sec (3x)\mid -\frac{1}{2}\ln \mid \sec (2x)\mid-\ln \mid \sec(x) \mid+\mathbb{C}$
Virgil Nicula
Mesaje: 244
Membru din: Sâm Oct 30, 2010 3:55 pm
Localitate: Bradenton, Florida

Re: O primitiva

Mesaj de Virgil Nicula »

Vom folosi identitatea conditionata $x+y+z=k\pi\iff \tan x+\tan y+\tan z=\tan x\tan y\tan z$ .

Asadar, $3x+(-2x)+(-x)=0\implies\tan 3x+\tan (-2x)+\tan (-x)=$ $\tan 3x\tan (-2x)\tan (-x)\implies$

$\tan 3x-\tan 2x-\tan x=\tan 3x\tan 2x\tan x$ etc etc
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