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

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- Mar 30th 2013, 04:23 AMsmokesalotIntegration
$\displaystyle \int tan(x)tan(2x)tan(3x)dx$

- Mar 30th 2013, 04:56 AMMINOANMANRe: Integration
Answer: (1/2)[ln(2(cosx)^2-1]+(2/3)ln(cosx)-(1/3)[4(sinx)^2-1] + Constant

- Mar 30th 2013, 06:02 AMsmokesalotRe: Integration
Actually, this is a prove question, I really need steps

thanks - Mar 30th 2013, 07:07 AMMINOANMANRe: Integration
I need to leave ....

But try tan(2x)=2tanx/(1-(tanx)^2) , tan(3x)=(3tanx-(tanx)^3)/(1-3(tanx)^2) - Mar 30th 2013, 12:54 PMMINOANMANRe: Integration
Smoke..

Just mow I came back home and I started to solve your integral...

I have discovered a very interesting result

believe it or not (tanx)(tan2x)(tan3x) = tan(3x)-tan(2x)-tanx

indeed :

tan(3x) = (tanx+tan(2x))/(1-tanxtan(2x))

this implies : tan(3x)(1-tanxtan(2x))=tanx+tan(2x) and after all simplifications we obtain : tan(3x)-tan(2x)-tan(x) = tanxtan(2x)tan(3x)

therefore your integral now is simple to calculate:

since integral(tanx)dx = -ln(cosx)

integral(tan(2x))dx = (-1/2)ln(cos(2x))

and Integral (tan(3x))dx = (-1/3)ln(cos(3x))

you can verify all these and check the calculations..

MINOAS - Apr 1st 2013, 01:40 AMsmokesalotRe: Integration
Thank you!

- Apr 2nd 2013, 11:39 AMMINOANMANRe: Integration
ok