InterviewSolution
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InterviewSolution
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`int(cosx)/(cos 3x)dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(cosx)/(cos 3x)dx` `=int(cosx)/(4cos^(3)x-3cosx)dx=int(dx)/(4cos^(2)x-3)` अंश व हर को `cos^(2)x` से भाग देने पर `therefore" "I=int(sec^(2)xdx)/(4-3sec^(2)x)` `=int(sec^(2)xdx)/(4-3(1+tan^(2)x))" "` (नोट कीजिए) `=int(sec^(2)xdx)/(1-3tan^(2)x)=int(sec^(2)xdx)/(1-(sqrt3tan x)^(2))` माना `" "sqrt3 tan x = t rArr sec^(2) x dx=(1)/(Sqrt3)dt` `therefore" "I=(1)/(sqrt3)int(dt)/(1-t^(2))=(1)/(sqrt3)int(dt)/((1-t)(1+t))` `=(1)/(2sqrt3)int{(1)/(1-t)+(1)/(1+t)}dt` `=(1)/(2sqrt3)log((1+t)/(1-t))` `=(1)/(2sqrt3)log((1+sqrt3 tan)/(1-sqrt3 tanx))` |
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