The Cardinal Sine Evaluation

\quad In\quad \int _{ \frac { 3 }{ \pi } }^{ \frac { 2 }{ \pi } }{ g\left( x \right) .dx } \quad \\ Let\quad x\quad =\quad f(t)\quad then\quad dx\quad =\quad f^{ ' }\left( t \right) .dt\\ Since\quad g\left( x \right) \quad is\quad inverse\quad of\quad f\left( x \right) ,\\ g\left( f\left( t \right) \right) \quad =\quad t\quad and\quad Using\quad property\quad \\ of\quad change\quad of\quad variable\quad the\quad ,\\ integral\quad becomes\quad :\\ \quad \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 2 } }{ t.f^{ ' }\left( t \right) .dt } \quad =\quad \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 2 } }{ x.f^{ ' }\left( x \right) .dx } \\ Now\quad by\quad using\quad integration\quad by\quad parts\quad ,\\ \int _{ \frac { 3 }{ \pi } }^{ \frac { 2 }{ \pi } }{ g\left( x \right) .dx } \quad =\quad \left[ x.f(x) \right] _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 2 } }\quad -\quad \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 2 } }{ f\left( x \right) .dx } \\ \\ \quad \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 2 } }{ f\left( x \right) .dx\quad +\quad \int _{ \frac { 3 }{ \pi } }^{ \frac { 2 }{ \pi } }{ g\left( x \right) .dx } } =\quad \sin { \frac { \pi }{ 2 } } -\sin { \frac { \pi }{ 6 } } \\ \\ Therefore\quad ,\quad Answer\quad =\quad 0.5\quad

$\text{The relation between f(x) and integral of its inverse g(x) is given by, } \\\large \displaystyle \int_{a}^{b} f(x) \, dx ~~+~~ \int_{c}^{d} g(x) \, dx =~~~~bd -ac.\\In~this~~~~~a=\dfrac \pi 6,~~~~~~b=\dfrac \pi 2,~~~~~~c=\dfrac 3 \pi ,~~~~d=\dfrac 2 \pi \\\therefore Int.= ~~ \dfrac \pi 2 * \dfrac 2 \pi ~~ -~~ \dfrac \pi 6 * \dfrac 3 \pi =~~~0.5. \\\text{This relation is true for any function and its inverse}\\ \text { not only for cardinal sine function .}$