`int(1)/(3+sin 2x)dx` का मान ज्ञात कीजिए ।

Explanation

माना `I=int(1)/(3+sin 2x)dx`
`=int(1)/(3+2sin x cosx)dx" "(because sin 2x = 2 sin x cos x)`
`=int(dx)/(3(cos^(2)x+sin^(2)x)+2sin x cos x)`
`" "[because 3 = 3.1=3(cos^(2)x +sin^(2)x)]`
`=int(sec^(2)x)/(3tan^(2)x+2 tan x+3)dx`
(अंश व हर को `cos^(2)x` से भाग देने पर))
माना `tan x = t rArr sec^(2)xdx=dt`
`therefore" "I=int(1)/(3t^(2)+2t+3)dt=(1)/(3)int(1)/(t^(2)+(2)/(3)t+1)dt`
`=(1)/(3)int(1)/((t+(1)/(3))^(2)+((2sqrt2)/(3))^(2))`
`=(1)/(3)int(1)/(t^(2)+(2)/(3)t+1)dt`
`=(1)/(3)int(1)/((t+(1)/(3))^(2)+((2sqrt2)/(3))^(2))`
`=(1)/(3).(1)/(((2sqrt2))/(3))tan^(-1)((t+(1)/(3))/((2sqrt2)/(3)))+c`
`=(1)/(2sqrt2)tan^(-1)((3t+1)/(2sqrt2))+c`
`=(1)/(2sqrt2)tan^(-1)((3 tan x+1)/(2sqrt2))+c`