Integrate( arcsinx arccos x)?

Hello,

∫ arcsinx arccosx dx =

let's consider that arccosx = (π/2) - arcsinx

(proof:

let arcsinx = t → sin t = x → (by the co-function identity sin t = cos[(π/2) - t]) → cos[(π/2) - t] = x;

if cos[(π/2) - t] = x, then (π/2) - t = arccosx → (substituting back arcsinx for t) → (π/2) - arcsinx = arccosx)

so the integral becomes:

∫ arcsinx [(π/2) - arcsinx] dx =

let:

arcsinx = t

hence:

x = sin t

dx = cos t dt

yielding, by substitution:

∫ arcsinx [(π/2) - arcsinx] dx = ∫ t [(π/2) - t] cos t dt =

∫ [(π/2)t - t²] cos t dt =

let's integrate by parts, letting:

(π/2)t - t² = u → [(π/2) - 2t] dt = du

cos t dt = dv → sin t = v

yielding:

∫ u dv = u v - ∫ v du

∫ [(π/2)t - t²] cos t dt = [(π/2)t - t²] sin t - ∫ sin t [(π/2) - 2t] dt =

(adjusting signs)

[(π/2)t - t²] sin t - ∫ [2t - (π/2)] (- sin t) dt =

let's integrate by parts further, letting:

2t - (π/2) = u → 2 dt = du

- sin t dt = dv → cos t = v

yielding:

∫ u dv = u v - ∫ v du

[(π/2)t - t²] sin t - {[2t - (π/2)] cos t - ∫ cos t 2 dt} =

(factoring the constant out)

[(π/2)t - t²] sin t - [2t - (π/2)] cos t + 2 ∫ cos t dt =

[(π/2)t - t²] sin t - [2t - (π/2)] cos t + 2sin t + C =

t [(π/2) - t] sin t - [2t - (π/2)] cos t + 2sin t + C =

let's now recall that:

t = arcsinx

sin t = x

hence:

cos t = √(1 - sin²t) = √(1 - x²)

then, substituting back:

t [(π/2) - t] sin t - [2t - (π/2)] cos t + 2sin t + C = arcsinx [(π/2) - arcsinx] x - [2arcsinx - (π/2)] √(1 - x²) + 2x + C =

arcsinx [(π/2) - arcsinx] x + [(π/2) - 2arcsinx] √(1 - x²) + 2x + C =

(splitting - 2arcsinx into - arcsinx - arcsinx)

arcsinx [(π/2) - arcsinx] x + [(π/2) - arcsinx - arcsinx] √(1 - x²) + 2x + C =

arcsinx [(π/2) - arcsinx] x + {[(π/2) - arcsinx] - arcsinx} √(1 - x²) + 2x + C =

(recalling that (π/2) - arcsinx = arccosx)

(arcsinx arccosx) x + (arccosx - arcsinx) √(1 - x²) + 2x + C =

ending with:

(arcsinx arccosx + 2) x + (arccosx - arcsinx)√(1 - x²) + C

I hope it's helpful