integration

**them** · April 7th 2006, 06:24 AM

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??

**TD!** · April 8th 2006, 06:21 AM

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.

$ \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 sec²(x), so:

$ \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):

$ \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:

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

So we conclude, without reduction-formula:

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

**them** · April 8th 2006, 11:54 AM

thanks, that was great help..i noticed the reduction formula gave effectively the same answer but in a different form.

**TD!** · April 8th 2006, 01:27 PM

Yes, that's perfectly possible and happens very often when you compare 'manual integration' and integration through such reduction formulas.

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