Intergation by parts - tricky trig question

using the substution x = cos u find the intergal of inverse cos x.
using the substitution x = tanx find the intergal of inverse tan x
help very much appreciated. text book says use given substitution then integrate by parts. thanks alot

1.
Integrate the original integrand by substitution:
∫ cosˉ¹x dx
Let x = cosu,
dx / du = -sinu
dx = -sinu du
u = cosˉ¹x
∫ cosˉ¹x dx = ∫ -usinu du

Integrate the new integrand by parts:
∫ -usinu du
Let f'(u) = -sinu
f(u) = cosu
Let g(u) = u
g'(u) = 1
∫ f'(u)g(u) du = f(u)g(u) - ∫ f(u)g'(u) du
∫ -usinu du = ucosu - ∫ cosu du
∫ -usinu du = ucosu - sinu

Put it all together to integrate the original function:
∫ cosˉ¹x dx = ∫ -usinu du
∫ cosˉ¹x dx = ucosu - sinu
Since u = cosˉ¹x,
∫ cosˉ¹x dx = (cosˉ¹x)cos(cosˉ¹x) - sin(cosˉ¹x) + C
∫ cosˉ¹x dx = xcosˉ¹x - √(1 - x²) + C

2.
Integrate the original integrand by substitution:
∫ tanˉ¹x dx
Let x = tanu,
dx / du = sec²u
dx = sec²u du
u = tanˉ¹x
∫ tanˉ¹x dx = ∫ usec²u du

Integrate the new integrand by parts:
∫ usec²u du
Let f'(u) = sec²u
f(u) = tanu
Let g(u) = u
g'(u) = 1
∫ f'(u)g(u) du = f(u)g(u) - ∫ f(u)g'(u) du
∫ usec²u du = utanu - ∫ tanu du
∫ usec²u du = utanu + ln|cosu| + C

Put it all together to integrate the original function:
∫ tanˉ¹x dx = ∫ usec²u du
∫ tanˉ¹x dx = utanu + ln|cosu| + C
Since u = tanˉ¹x,
∫ tanˉ¹x dx = tanˉ¹xtan(tanˉ¹x) + ln[cos(tanˉ¹x)] + C
∫ tanˉ¹x dx = xtanˉ¹x + ln[1 / √(x² + 1)] + C
∫ tanˉ¹x dx = xtanˉ¹x - ln√(x² + 1) + C
∫ tanˉ¹x dx = xtanˉ¹x - ln(x² + 1) / 2 + C
∫ tanˉ¹x dx = [2xtanˉ¹x - ln(x² + 1)] / 2 + C