1.

Evaluate : (i) `intsin^(-1)(3x-4x^(3))dx` (ii) `intsin^(-1)((2x)/(1+x^(2)))dx` (iii) `inttan^(-1)sqrt((1-x)/(1+x))dx` (iv) `intsin^(-1)sqrt((x)/(a+x))dx`

Answer» (i) Put x=sin t so that dx = cos t dt.
`:.intsin^(-1)(2x-4x^(3))dx=intsin^(-1)(3sintsin^(3)t)costdt`
`=intsin^(-1)(3sin3t)costdt`
`=3inttcostdt`
`=3[t(sint)-int*sintdt]" [integrating by parts]"`
`=3tsint+3cost+C`
`=3x(sin^(-1)x)+3sqrt(1-x^(2)+C)`.
(ii) Put x=tan t so that `=sec^(2)tdt`.
`:.intsin^(-1)((2x)/(1+x^(2)))dx=intsin^(-1)((2tant)/(1+tan^(2)t))sec^(2)tdt`
`=intsin^(-1)(sin2t)sec^(2)tdt=2intt*sec^(2)tdt`
`=2t*tant+2log|cost|+C`
`2x(tan^(-1)x)+2log|{:(1)/(sqrt(1+x^(2))):}|+C`
`=2x(tan^(-1)x)+2*(-(1)/(2))log|1+x^(2)|+C`
`=2x(tan^(-1)x)-log|{:1+x^(2):}|+C`.
(iii) Put x=cost so that dx =-sin t dt.
`:,inttan^(-1)sqrt((1-x)/(1+x))dx=inttan^(-1)sqrt((1-cost)/(1+cost))(-sint)dt`
`=inttan^(-1)sqrt((2sin^(2)(t//2))/(2cos^(2)(t//2)))(-sint)dt`
`int{:[tan^(-1)("tan"(t)/(2))]:}(-sint)dt=-(1)/(2)int(sint)dt`
`=-(1)/(2)[t(-cost)-int1*(-cos)dt]" [integrating by parts]"`
`=(1)/(2)t*cos-(1)/(2)sint+C=(1)/(2)x(cos^(-1)x)-(1)/(2)sqrt(1-x^(2))+C`.
(iv) Put `x=atan^(2)t` so that dx `=(2asec^(2)ttant)` dt.
`:.intsin^(-1)sqrt((x)/(a+x))dx=intsin^(-1){sqrt((atan^(2)t)/(a(1+tan^(2)t)))}2asec^(2)t tantdt`
`=2aint t(sec^(2)t*tan t)dt`
`=2a{:[t*(1)/(2)tan^(2)t-int1*(1)/(2)tan^(2)tdt]:}`
[integrating by parts and using `intsec^(2)t tantdt=(1)/(2)tan^(2)t]`
`=at(tan^(2)t)-aint(sec^(2)t-1)dt`
`at(tan^(2)t)-aintsec^(2)t+aintdt`
`at(tan^(2)t)-atant+at+C`
`=xtan^(-1)sqrt((x)/(a))-sqrt(ax)+atan^(-1)sqrt((x)/(a))+C`.


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