# Evaluating integral!

• Mar 10th 2009, 11:25 PM
twilightstr
Evaluating integral!
evaluate the integral of 1/cosx(sinx)^5
• Mar 11th 2009, 04:50 AM
mr fantastic
Quote:

Originally Posted by twilightstr
evaluate the integral of 1/cosx(sinx)^5

Make the substitution $u = \sin x$. Then the integral becomes $\int \frac{du}{(1 - u^2) u^5}$.
• Mar 11th 2009, 11:59 PM
twilightstr
however,
cosx is in the denominator
• Mar 12th 2009, 01:04 AM
Quote:

Originally Posted by twilightstr
however,
cosx is in the denominator

Yes but there is also another cos(x)

sin(x) =u

cos(x)dx = du
dx= du/cos(x)

Put this in integral
$
\int{\frac{du}{cos(x) \cdot cos(x) \cdot sin^5(x) }}$

$\int{\frac{du}{(1-u^2) \cdot u^5}}$
• Mar 12th 2009, 01:42 AM
simplependulum
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 AM
chisigma
Under the assumption that the function we have to integrate is…

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

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

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

… so that...

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

... 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