hi, is there any method of integrating a hyperbolic or trignometric function whos degree is 4, such as (tanx)^4, (cosechx)^4 WITHOUT resorting to reduction formulae??

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- Apr 7th 2006, 06:24 AMthemintegration
hi, is there any method of integrating a hyperbolic or trignometric function whos degree is 4, such as (tanx)^4, (cosechx)^4 WITHOUT resorting to reduction formulae??

- Apr 8th 2006, 06:21 AMTD!
Sure, there just isn't a "golden recipe" which will be succesfull every time. You'll need to play arround with trig formulas. I'll do one as an example.

$\displaystyle

\int {\tan ^4 x} dx = \int {\tan ^2 x\tan ^2 x} dx = \int {\tan ^2 x\left( {\sec ^2 x - 1} \right)} dx

$

Now I'll split the integral in two. The first one is easy since the derivative of tan(x) is exactly secē(x), so:

$\displaystyle

\int {\tan ^2 x\sec ^2 x} dx = \int {\tan ^2 x} d\left( {\tan x} \right) = \frac{{\tan ^3 x}}{3} + C

$

For the second one; I convert to sin(x) and cos(x):

$\displaystyle

\int { - \tan ^2 x} dx = - \int {\frac{{\sin ^2 x}}{{\cos ^2 x}}} dx = \int {\frac{{\cos ^2 x - 1}}{{\cos ^2 x}}} dx

$

Splitting the integral in two again gives us simply:

$\displaystyle

\int {dx} - \int {\frac{1}{{\cos ^2 x}}} dx = x - \tan x + C

$

So we conclude, without reduction-formula:

$\displaystyle

\int {\tan ^4 x} dx = \frac{{\tan ^3 x}}{3} - \tan x + x + C

$ - Apr 8th 2006, 11:54 AMthemreply
thanks, that was great help..i noticed the reduction formula gave effectively the same answer but in a different form.

- Apr 8th 2006, 01:27 PMTD!
Yes, that's perfectly possible and happens very often when you compare 'manual integration' and integration through such reduction formulas.