How do you find the integral of $(5 csc x d x)$ $(5 csc x dx)$ from $[\frac{π}{2}, π]$ $[pi/2,pi]$ ?

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Answer 1 · 2015-03-08T14:33:23

$\int_{\frac{π}{2}}^{π} 5 csc x d x$ $int_(pi/2)^pi 5cscx dx$

This is an improper integral, and it diverges (to $\infty$ $oo$ ).
(It does not converge.)

Solution:
The integral is improper because $π$ $pi$ is not in the domain of $csc$ $csc$ , so we need :

$lim_(brarrpi^-)int_(pi/2)^b 5cscx dx=5lim_(brarrpi^-)int_(pi/2)^b cscx dx$
(It's hard to see, but both limits are $brarrpi^-$ ,)

We'll need to find $intcscxdx$ .

$intcscxdx=int1/sinx dx=int1/(2sin(x/2)cos(x/2)) dx$

$=1/2int(sin^2(x/2)+cos^2(x/2))/(sin(x/2)cos(x/2))dx$

$=1/2int(sin(x/2)/cos(x/2)+cos(x/2)/sin(x/2))dx$

$=1/2[-2ln(cos(x/2)+2lnsin(x/2)]+C$

$=lnsin(x/2)-lncos(x/2)+C$ $=lntan(x/2)+C$

So the definite integral is:
, $int_(pi/2)^b cscx dx=lntan(b/2)-lntan(pi/4)=lntan(b/2)-ln1=lntan(b/2)$

Finally, as $brarr pi ^-$ , we see that $b/2rarr (pi/2)^-$ and $tan(b/2)rarroo$ , so as $brarr pi ^-$ , we get $lntan(b/2)rarroo$

$lim_(brarrpi^-)int_(pi/2)^b 5cscx dx=5lim_(brarrpi^-)int_(pi/2)^b cscx dx=5(oo)=oo$ .

The integral diverges.

How do you find the integral of (5 csc x dx)(5cscxdx)(5 csc x dx) from [pi/2,pi][π2,π][pi/2,pi]?