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(1)/(2-3sin2x)...

`(1)/(2-3sin2x)`

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`"Let I"=int (1)/(2-3sin 2x)dx`
`"Put tan x"=1 therefore x=tan^(-1)t`
`therefore dx=(dt)/(1+t^(2)) and sin2x=(2t)/(1+t^(2))`
`therefore I=int (1)/(2-3((2t)/(1+t^(2)))).(dt)/(1+t^2)`
`=int (1+t^(2))/(2+2t^(2)-6t).(dt)/(1+t^2)`
`=int (1)/(2t^2-6t+2)dt`
`=(1)/(2)int (1)/((t^2-3t+(9)/(4))-(9)/(4)+1)dt`
`=(1)/(2)int (1)/((t-(3)/(2))-((sqrt5)/(2))^(2))dt`
`=(1)/(2)xx(1)/(2xx(sqrt5)/(2))log|(t-(3)/(2)-(sqrt5)/(2))/(t-(3)/(2)+(sqrt5)/(2))|+c`
`=(1)/(2sqrt5)log|(2t-3-sqrt5)/(2t-3+sqrt5)|+c`
`=(1)/(2sqrt5)log|(2tan-3-sqrt5)/(2tan x-3+sqrt5)|+c`
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