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(sin x)/(sin 3x)...

`(sin x)/(sin 3x)`

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`"Let I"=int (sin x)/(sin 3x)dx=int (sin x)/(3 sin x-4sin^(2)x)dx=int (1)/((3-4)sin^(2)x)dt`
Dividing both numberator and denominator by `cos^(2)x`, we get
`I=int (sec^(2)x)/(3sec^(2)x-4tan^(2)x)dx`
`I=int (sec^(2)x)/(3-tan^(2)x)dx`
Put tan x=t `therefore sec^(2)xdx=dt`
`therefore I=int (1)/(3-t^(2))dt=int (1)/((sqrt3)^(2)-t^(2))dt`
`=(1)/(2sqrt3)log |(sqrt3+t)/(sqrt3-t)|+c`
`=(1)/(2sqrt3)log|(sqrt3+tanx)/(sqrt3-tanx)|+c`
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