Integration-Cauchy

Y use some thm....use ur common sense....imagine the graph in ur head...u have

1/cosx sin^2 x

from 0 to pi/2 u have +ve values..from pi/2 to pi u have negative values but equal in value.......they simply cancel of as I-I=0

OR

f(0+x)=-f(pi-x) to be precise..

$\int _{ 0 }^{ \pi }{ \frac { 1 }{ \cos { x } (1-\cos ^{ 2 }{ x } ) } dx } \\ =\quad \int _{ 0 }^{ \pi }{ \frac { \cos { x } }{ \cos ^{ 2 }{ x } (1-\cos ^{ 2 }{ x } ) } dx } \\ =\quad \int _{ 0 }^{ \pi }{ \frac { \cos { x } }{ (1-\sin ^{ 2 }{ x } )(\sin ^{ 2 }{ x } ) } dx } \\ =\quad \int _{ 0 }^{ 0 }{ \frac { 1 }{ (1-{ t }^{ 2 })({ t }^{ 2 }) } dt } \quad \quad [t\quad =\quad \sin { x } ]\\ using\quad Cauchy\quad principal\quad value\quad \\ =\quad \boxed { 0 }$

$Well\quad it\quad can\quad also\quad be\quad done\quad by\quad putting\quad \\ the\quad numerator\quad as\\ \quad \\ 1\quad =\quad \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } \\ \\ Then\quad the\quad integral\quad will\quad become\\ \\ I\quad =\quad \int _{ 0 }^{ \pi }{ \frac { \cos ^{ 2 }{ x } +\sin ^{ 2 }{ x } }{ \cos { x } .\sin ^{ 2 }{ x } } .dx } \\ I\quad =\quad \int _{ 0 }^{ \pi }{ (\cot { x } .\csc { x } \quad +\quad \sec { x } } ).dx\quad \quad \quad -\quad equation\quad 1.....\\ Now\quad using\quad property\\ \int _{ a }^{ b }{ f\left( x \right) =\int _{ a }^{ b }{ f(a+b-x) } } .dx\\ we\quad get\\ I\quad =\quad -\quad (\int _{ 0 }^{ \pi }{ (cotx.cscx\quad +\quad secx).dx } )\quad \quad -\quad equation\quad 2.....\\ Now\quad by\quad adding\quad the\quad two\quad equations\quad we\quad get\\ \quad \quad \quad \quad \quad 2I\quad =\quad 0\\ Therefore\quad I\quad =\quad 0\\$

In fact the answer should be undefined.

We have: $\int_0^\pi \frac1{\cos x(1-\cos^2x)}dx$

Using the famous identity that $\cos^2x+\sin^2x=1$ : $=\int_0^\pi \frac1{\cos x\sin^2x} dx$

Since we want to make the denominator have as less power as possible, we raise the degree of the numerator: $=\int_0^\pi \frac{\cos^2x+\sin^2x}{\cos x\sin^2x}dx$

Break it open and simplify: $=\int_0^\pi \frac{\cos x}{\sin^2x}dx + \int_0^\pi \frac1{\cos x}dx$

Converting to non-fraction: $=\int_0^\pi\cot x\csc xdx+\int_0^\pi\sec xdx$

Using formulae: $=[\csc x]_0^\pi+[\ln|\sec x+\tan x|]_0^\pi$

For the first term, the value at the bounds are undefined!