The value of `int(1)/(3 sin x -cos x +3)dx` is equal to

Options

A. ` log (( tan "" ( x ) /( 2 ) + 1 ) /( 2 tan "" ( x ) /( 2) + 1 ) ) +C `
B. ` ( 1 ) /( 2 ) log ( ( 2tan "" ( x ) /( 2 ) + 1 ) / ( tan "" ( x ) /(2 ) +1 ) ) + C `
C. `log ( ( 2tan "" ( x ) /( 2 ) + 1 ) /( tan "" ( x ) /( 2 ) + 1 ) ) + C `
D. ` 2log (( 2 tan "" ( x ) /( 2 ) + 1 ) /( tan "" ( x ) /( 2) + 1 ) ) +C `

Answer

C

Explanation

Let I ` = int (dx )/ ( 3 sin x - cos x + 3 ) `
` {{:( because, sin x = (2 tan "" ( x ) / ( 2 )) /( 1 + tan ^ 2 "" ( x )/ ( 2 )) ) , ( and , cos x = ( 1 - tan ^ 2 "" ( x )/(2 ))/(1 +tan ^( 2) "" ( x )/ ( 2 ))):}} `
` I = int ( dx ) / ( 3 { ( 2 tan "" (x ) / ( 2 ) ) / ( 1 + tan ^ 2 "" ( x )/ ( 2 ) )} - { ( 1 - tan ^ 2 "" ( x ) / ( 2 ))/ ( 1 + tan ^ 2"" ( x )/ ( 2 ))} + 3 ) `
` = int (( 1 + tan ^ 2 "" ( x )/ ( 2 )) dx )/ ( 6 tan "" ( x )/ ( 2 ) - 1 + tan ^ 2 "" ( x ) / ( 2 ) + 3 + 3tan ^( 2 ) "" ( x ) / ( 2 ) ) `
` = int (sec^ 2 "" ( x ) / ( 2 ) ) / ( 4 tan ^ 2 "" (x ) / ( 2 ) + 6 tan"" (x )/ (2 ) + 2) dx `
( let ` t = tan "" ( x )/ ( 2 ), dt= ( 1 )/ (2) sec^ 2 "" ( x )/ (2 ) dx` )
` = int ( dt ) / ( 2 t^ 2 + 3t + 1 ) `
` = int ( dt )/ ( ( t + 1 ) ( 2t+ 1 ) ) `
` = int { ( - 1 )/ ( ( t + 1 ) ) + ( 2 ) /( ( 2t + 1 ))} dt ` ( by partical fraction )
` = -log ( t + 1 ) + ( 2) / ( 2 ) log ( 2 + 1 ) + C `
` = log"" (( 2t + 1 ))/( (t + 1 )) + C `
` = log (( 2 tan"" ( x )/ ( 2 ) + 1 )/ ( tan "" ( x ) / ( 2 ) + 1 ) ) + C `