Question #e9266

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Answer 1 · 2017-08-29T16:06:46

After using u= $cot^(-1)(sqrt(1+x)/sqrt(1-x))$ transform,

$cotu=sqrt(1+x)/sqrt(1-x)$

$(cotu)^2=(1+x)/(1-x)$

$(1-x)*(cotu)^2=1+x$

$(cotu)^2-x*(cotu)^2=1+x$

$x*[(cotu)^2+1]=(cotu)^2-1$

$x*(cscu)^2=(cscu)^2*[(cosu)^2-(sinu)^2]$

$x=(cosu)^2-(sinu)^2$

$x=cos2u$

Hence,

$y=(sin[cot^(-1)(sqrt(1+x)/sqrt(1-x))])^2$

= $(sinu)^2$

= $1/2*2(sinu)^2$

= $1/2*(1-cos2u)$

= $(1-x)/2$

1) I used $u=cot^(-1)(sqrt(1+x)/sqrt(1-x))$ transform for finding x in terms of u. Finally I found $x=cos2u$ .

2) I rewrote $(sinu)^2$ in terms of $cos2u$ . Finally I found $(sin[cot^(-1)(sqrt(1+x)/sqrt(1-x))])^2$ as $(1-x)/2$ .