1.

Evaluate : (i) intcos^(-1)xdx (ii) inttan^(-1)xdx (iii) intsec^(-1)xdx

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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