∫arcsec(√x) dx
integrate by parts
let u = arcsec(√x) : dv = dx
du = (1/2) (1/√x) dx /√(x) √(1 - x) = (1/2) dx /x √(1-x) : v = x
∫u dv = uv - ∫ v du
∫arcsec(√x) dx = x arcsec(√x) - 1/2 ∫ x dx / x √(1-x)
=x arcsec(√x) - 1/2 ∫ dx / √(1-x)
let 1- x = y
-dx = dy
= x arcsec(√x) - 1/2 ∫ -dy / √y
= x arcsec(√x) +(1/2) √y /(1/2)
= x arcsec(√x) + √y + c
substitute back y = (1-x)
= x arcsec(√x) + √(1-x) + c
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Verified answer
∫arcsec(√x) dx
integrate by parts
let u = arcsec(√x) : dv = dx
du = (1/2) (1/√x) dx /√(x) √(1 - x) = (1/2) dx /x √(1-x) : v = x
∫u dv = uv - ∫ v du
∫arcsec(√x) dx = x arcsec(√x) - 1/2 ∫ x dx / x √(1-x)
=x arcsec(√x) - 1/2 ∫ dx / √(1-x)
let 1- x = y
-dx = dy
= x arcsec(√x) - 1/2 ∫ -dy / √y
= x arcsec(√x) +(1/2) √y /(1/2)
= x arcsec(√x) + √y + c
substitute back y = (1-x)
= x arcsec(√x) + √(1-x) + c