`int1/(cosx+sqrt(3)sinx)` dx equalsA. `logtan((pi)/(2)+(pi)/(12))+C`B. `logtan((x)/(2)-(pi)/(12))+C`C. `(1)/(2)log tan ((x)/(2)+(pi)/(12))+C`D. `(1)/(2)log tan((x)/(2)-(pi)/(12))+C`

`int1/(cosx+sqrt(3)sinx)` dx equalsA. `logtan((pi)/(2)+(pi)/(12))+C`B.

1.	`int1/(cosx+sqrt(3)sinx)` dx equalsA. `logtan((pi)/(2)+(pi)/(12))+C`B. `logtan((x)/(2)-(pi)/(12))+C`C. `(1)/(2)log tan ((x)/(2)+(pi)/(12))+C`D. `(1)/(2)log tan((x)/(2)-(pi)/(12))+C`
Answer» Correct Answer - c We have , `I=int(1)/(cosx +sqrt(3)sin x)dx` `rArr I= (1)/(2)int(1)/((1)/(2)cosx+(sqrt(3))/(2)sinx)dx` `rArrI=(1)/(2)int (1)/(cos x "cos" (pi)/(3)+sin x " sin" (pi)/(3))dx` `rArrI=(1)/(2)int(1)/(cos(x-(pi)/(3)))dx = (1)/(2) int sec(x-(pi)/(3))dx` `rArr I= (1)/(2)log tan ((pi)/(4)+(pi)/(2)-(pi)/(6))+C = (1)/(2) log tan((x)/(2)+(pi)/(12))+C`

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`int1/(cosx+sqrt(3)sinx)` dx equalsA. `logtan((pi)/(2)+(pi)/(12))+C`B. `logtan((x)/(2)-(pi)/(12))+C`C. `(1)/(2)log tan ((x)/(2)+(pi)/(12))+C`D. `(1)/(2)log tan((x)/(2)-(pi)/(12))+C`

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