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