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#