InterviewSolution
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`int(sinx)/((1+cosx)(2+3cosx))dx` का मान ज्ञात कीजिए । |
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Answer» माना `I=int(sinx)/((1+cosx)(2+3cosx))dx` `=int(-dt)/((1+t)(2+3t))` `" माना "cosx=t` `" "-sinx=(dt)/(dx)` `" "rArr sinx dx =-dt` माना `(-1)/((1+t)(2+3t))=(A)/(1+t)+(B)/(2+3t)=(A(2+3t)+B(1+t))/((1+t)(2+3t))` `rArr" "A(2+t)+B(1+t)=-1` `t=-1` रखने पर `A(2-3)+0=-1` `rArr" "A=1` `t=-(2)/(3)` रखने पर `0+B(1-(2)/(3))=-1` `rArr" "B=-3` `therefore" "I=int((1)/(1+t)-(3)/(2+3t))dt` `=log|1+t|-log|2+3t|+c` `=log|(1+t)/(2+3t)|+c` `=log|(1+cosx)/(2+3cosx)|+c` |
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