since cos(x) dx = d ( sin(x) ), upon substituting sin(x) = u your integral is
int ( 1/sqrt( 1+u^2) du ) u= 0..1
Now substitute u = sinh(v), then du = cosh(v) dv, sqrt(1+u^2) = cosh(v),
so the integral just becomes int ( dv) , v = 0..arsinh(1) = arsinh(1)
Note that for the inverses of hyperbolic functions you should write as a prefix ar, not arc. So it is arsinh, not arcsinh because arcs have nothing to do with hyperbolic functions unlike sin and cos.
Comments
since cos(x) dx = d ( sin(x) ), upon substituting sin(x) = u your integral is
int ( 1/sqrt( 1+u^2) du ) u= 0..1
Now substitute u = sinh(v), then du = cosh(v) dv, sqrt(1+u^2) = cosh(v),
so the integral just becomes int ( dv) , v = 0..arsinh(1) = arsinh(1)
Note that for the inverses of hyperbolic functions you should write as a prefix ar, not arc. So it is arsinh, not arcsinh because arcs have nothing to do with hyperbolic functions unlike sin and cos.
int(cos(x)/sqrt(1+sin(x)^2), x = 0 .. (1/2)*Pi)
=
ArcSinh(1)
=
0.881374