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Evaluate : (i) intcos^(-1)xdx (ii) inttan^(-1)xdx (iii) intsec^(-1)xdx |
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Answer» SOLUTION :(i) Put `COS^(-1)x=t" so that "x=cost anddx=-sintdt`. `:.intcos^(-1)xdx=-inttsintdt` `=-[t*(-cost)-int*(-cost)DT]` [integrating by parts] `=tcost-intcostdt=tcost-sint+C` `=XCOS^(-1)x-sqrt(1-x^(2))+C` `[becausecost=xrArrsint=sqrt(1-x^(2))]`. (II) Put `tan^(-1)x=t" so that "x=tant anddx=sec^(2)tdt`. `:.inttan^(-1)xdx=inttsec^(2)tdt` `=t*tant-int1*tantdt""` [integrating by parts] `=t*tant+log|cost|+C` `=(tan^(-1)x)*x+log|(1)/(sqrt(1+x^(2)))|+C` `[becausetant=xrArrcost=(1)/(sqrt(1+x^(2)))]` `=x(tan^(-1)x)-(1)/(2)log|1+x^(2)|+C`. (iii) Put `sec^(-1)x=t" so that"x=sectanddx=sec t tantdt`. `:.intsec^(-1)xdx=intt(sec t tan t)dt` `=t(sect)-int1*sect dt""` [integrating by parts] `=t(sect)-log|sec t+tant|+C` `=t(sect)-log|sect+sqrt(sec^(2)t-1)|+C` `=x(sec^(-1)x)-log|x+sqrt(x^(2)-1)|+C`. |
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