1.

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

Answer» (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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