evaluate the integral of 1/cosx(sinx)^5

Printable View

- Mar 10th 2009, 11:25 PMtwilightstrEvaluating integral!
evaluate the integral of 1/cosx(sinx)^5

- Mar 11th 2009, 04:50 AMmr fantastic
- Mar 11th 2009, 11:59 PMtwilightstr
however,

cosx is in the denominator - Mar 12th 2009, 01:04 AMADARSH
- Mar 12th 2009, 01:42 AMsimplependulum
1/(1-u^2)(u^5)

= (1/u)[ 1/(1-u^2)(u^4) ]

= (1/u)[ (1+u^2)/(u^4) + 1/(1-u^2) ]

= (1+u^2)/(u^5) + 1/(1-u^2)(u)

= 1/u^5 + 1/u^3 + 1/u + u/(1-u^2) - Mar 12th 2009, 02:52 AMchisigma
Under the assumption that the function we have to integrate is…

$\displaystyle f(x)= \frac {1}{\cos x \cdot sin^{5} x}$

… the ‘standard approach’ is the change of variable…

$\displaystyle t=\tan \frac {x}{2}$

… so that...

$\displaystyle \sin x = \frac {2t}{1+t^{2}}, \cos x= \frac {1-t^{2}}{1+t^{2}}, dx= 2 \cdot \frac {dt}{1+t^{2}}$

... and we arrive to an integral that contains rational functions and that can be solved without great difficulty. Other substitutions doesn’t work so well … at least in my opinion…

Regards