Evaluate:`int1/(3+sin2x)dx`

1.	Evaluate:`int1/(3+sin2x)dx`
Answer» Correct Answer - `(1)/(sqrt(8)) "tan"^(-1) (3tanx+1)/(sqrt(8))+c` `I=int(dx)/(3+sin2x)` `=int(dx)/(3+sinx cosx)` ` =int(sec^(2)xdx)/(3sec^(2)x+2tanx)" "("Dividing Nr. and Dr. by " cos^(2)x)` ` =int(sec^(2)xdx)/(3tan^(2)x+2tanx+3) ` ` =int(dt)/(3t^(2)+2t+3)" " ("Putting " tanx=t)` `=(1)/(3) int (dt)/(t^(2)+(2)/(3)t+1)` ` = (1)/(3) int (dt)/((t+(1)/(3))^(2)+(8)/(9))` ` = (1)/(3)* (1)/(sqrt(8)//3) "tan"^(-1) ((t+(1)/(3)))/((sqrt(8))/(3))+c ` ` =(1)/(sqrt(8)) "tan"^(-1) (3tanx+1)/(sqrt(8))+c`

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