Question #f628c

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Answer 1 · 2016-10-29T14:09:28

Given expression $=arcsin((x+1)/sqrt(2*(x²+1)))$
Let $x = tantheta$

So $theta=tan^-1x$

Inserting $x = tantheta$ the given expression becomes

$=arcsin((tantheta+1)/sqrt(2*(tan^2theta+1)))$

$=arcsin((sintheta/costheta+1)/sqrt(2*(sec^2theta)))$

$=arcsin(1/sqrt2((sinthetasectheta+1)/(sectheta)))$

$=arcsin(1/sqrt2((sinthetacancelsectheta)/cancelsectheta+1/(sectheta)))$

$=arcsin(1/sqrt2(sintheta+costheta))$

$=arcsin(1/sqrt2sintheta+1/sqrt2costheta)$

$=arcsin(cos(pi/4)sintheta+sin(pi/4)costheta)$

$=arcsin(sin(theta+pi/4))$

$=theta+pi/4$

$=tan^-1x+pi/4$