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Question: integrate (arcsin x dx)

I = int arcsin x dx

Let u = arcsin(x)

Then x = sin(u)

And, dx = cos(u) du

This gives us,

I = int u.cos(u) du

Doing integration by parts,

I = u.sin(u) - int sin(u) du

I = u.sin(u) + cos(u)

I = arcsin(x).x + √(1 - sin^2(u))

I = x.arcsin(x) + √(1 - x^2)

by Level 11 User (81.5k points)

Given arc sinx dx

let.......dv/dx = 1

∫dv=∫dx

v=x,

and ... u=arc sinx

du/dx= 1/(1 - x^2)

arcsinx=x.arcsinx - x(1 - x^2)^-1/2 dx

But......∫x(1 - x^2)^-1/2 dx = 

let.......t=1 - x^2.......... dt/dx = -2x............... dx = -dt/2x

=∫xt^-1/2.(-dt/2x)....... = -t^1/2............... = -(1 - x^2)^1/2

therefore

∫arcsinx=x.arcsinx - { - (1 - x^2)}

=x.arcsinx + √(1 - x^2)

by Level 3 User (4.0k points)

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