Question
Find the integral: $ \int \frac { \sec ^ { 2 } x } { cosec ^ { 2 } x } d x$

Answer

Let I = $\int \frac { \sec ^ { 2 } x } { cosec ^ { 2 } x } d x = \int \frac { \left( \frac { 1 } { \cos ^ { 2 } x } \right) } { \left( \frac { 1 } { \sin ^ { 2 } x } \right) } d x$
$= \int \frac { \sin ^ { 2 } x } { \cos ^ { 2 } x } d x$
$ = \int \tan ^ { 2 } x d x = \int \left( \sec ^ { 2 } x - 1 \right) d x$ $ \left[ \because \tan ^ { 2 } x = \sec ^ { 2 } x - 1 \right]$
$ = \int \sec ^ { 2 } x d x - \int 1 d x = \tan x - x + c$

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