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InterviewSolution
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`intsqrt(cotx)dx` का मान ज्ञात कीजिए । |
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Answer» `int sqrt(cotx)dx` माना `cot x = t^(2)` तथा `-"cosec"^(2)xdx=2tdt rArr dx=(-2t)/((1+t^(4)))dt,` तब `intsqrt(cot x)dx=-int(2t^(2))/((t^(4)+1))dt` `=-int([(t^(2)+1)+(t^(2)-1)])/((t^(4)+1))dt` `-int((t^(2)+1))/((t^(4)+1))dt-int((t^(2)-1))/((t^(4)+1))dt` `=-int((1+(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt-int((1-(1)/(t^(2))))/((t^(2)+(1)/(t^(2))))dt` `=-int((1+(1)/(t^(2))))/([(t-(1)/(t))^(2)+2])dt-int((1-(1)/(t^(2))))/([(t+(1)/(t))^(2)-2])dt` अब माना `(t-(1)/(t))= u rArr (1+(1)/(t^(2)))dt=du` तथा `t+(1)/(t)= v rArr (1-(1)/(t^(2)))dt=dv,` तब `=int(du)/([u^(2)+(sqrt2)^(2)])-int(dv)/([v^(2)-(sqrt2)^(2)])` `=-(1)/(sqrt2)tan^(-1)((u)/(sqrt2))-(1)/(2sqrt2)log((v-sqrt2)/(v+sqrt2))+c` `=-(1)/(sqrt2)tan^(-1)((t-(1)/(t))/(sqrt2))-(1)/(2sqrt2)log((t+(1)/(t)-sqrt2)/(t+(1)/(t)+sqrt2))+c` `=-(1)/(sqrt2)tan^(-1)((t^(2)-1)/(sqrt2t))-(1)/(2sqrt2)log[(t^(2)-sqrt2t+1)/(t^(2)+sqrt2t+1)]+c` `=-(1)/(sqrt2)tan^(-1)((cot x-1)/(sqrt(2cosx)))-(1)/(2sqrt2)log[(cosx-sqrt(2cosx)+1)/(cotx-sqrt(2cotx)+1)]+c` |
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