Find the indefinite integral of ∫ √ (a ^ 2-x ^ 2) DX,

Find the indefinite integral of ∫ √ (a ^ 2-x ^ 2) DX,

Let x = asiny, DX = acosy dy
√(a²-x²)=√(a²-a²sin²y)=acosy
The original formula = A & sup2; ∫ cos & sup2; y dy
=a²/2*∫(1+cos2y) dy
=a²/2*y+a²/2*1/2*∫cos2y d(2y)
=a²/2*y+a²/4*sin2y
=a²/2*y+a²/2*sinycosy
=a²/2*arcsin(x/a)+a²/2*(x/a)*√(a²-x²)/a
=(a²/2)arcsin(x/2)+(x/2)√(a²-x²)+C

Finding indefinite integral ∫ X / (x ^ 2) DX

Solution
∫x/(x^2)dx
=∫1/xdx
=ln|x|+C

Find indefinite integral ∫ x ^ 2 / √ (a ^ 2-x ^ 2) DX =?
integration by parts
∫x^2/√(a^2-x^2)dx=x^2*arcsin(x/a)-∫2x/√(a^2-x^2)dx=x^2*arcsin(x/a)-2xarcsin(x/a)+2arcsin(x/a)
=(x^2-2x+2)arcsin(x/a)+C
What's wrong with that?

Let x = Asian, then DX = acost DT 8747\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\2tdt = A & # 178; t / 2-1 / 2. A & # 178; sin2t + C = 1 / 2. A & # 178; arcsin (x / a) - x · √ (A & # 178; - X & # 178;) + C