Question

Integration dx/√2ax-x^2=a^n sin^-1[x/a-1]

Answer 1

Given ∫dx/√(2ax-x2)

= ∫dx/√(2ax-x2 - a2 + a2 )

= ∫dx /√(a2 - (a2 - 2ax + x2)

= ∫dx /√((a2 - (x - a)2) = sin-1 [(x-a) / a]

Now comparing an = 1 when n = 0 .

Let x = 0 LHS is value is 0 then RHS = sin-1 (0) = 0.

Answer 2

Answer:

Step-by-step explanation:

Given ∫dx/√(2ax-x2)

= ∫dx/√(2ax-x2 - a2 + a2 )

= ∫dx /√(a2 - (a2 - 2ax + x2)

= ∫dx /√((a2 - (x - a)2) = sin-1 [(x-a) / a]

Now comparing an = 1 when n = 0 .

Let x = 0 LHS is value is 0 then RHS = sin-1 (0) = 0.