InterviewSolution
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InterviewSolution
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`int(sqrt(tanx)+sqrt(cotx))dx` का मान ज्ञात कीजिए । |
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Answer» `int(sqrt(tanx+sqrt(cotx)))dx` `=int(sqrt(tanx)+(1)/(sqrt(tanx)))dx=int((tanx+1)/(sqrt(tanx)))dx` माना `x=t^(2) rArr x = tan^(-1) t^(2)` व `dx=(2t)/((1+t^(4)))dt` `=int((t^(2)+1))/(t).(2t)/((1+t^(4)))dt=2int((t^(2)+1))/((t^(2)+1))dt` `=2int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt=2int((1+(1)/(t^(2))))/((t-(1)/(t))^(2)+2)dt` माना `(t-(1)/(t))=u rArr (1+(1)/(t^(2)))dt=du,` तब `2int((1+(t)/(t^(2))))/((t-(1)/(t))^(2)+2)dt=2int(du)/((u^(2)+2))` `=2.(1)/(sqrt2)tan^(-1).(u)/(sqrt2)+c` `=sqrt2 tan^(-1).((t-(1)/(t)))/(sqrt2)+c` `=sqrt2 tan^(-1).((t^(2)-1))/((sqrt2t))+c` `=sqrt2 tan^(-1)((tanx-1)/(sqrt2 tanx))+c` |
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