Inverse trigonometry quetion.....plzz can anybody solve it

sqrt{ a-x / a+x} dx limit is from -a to a

√ { (a-x) / (a+x) } multiple both sides with √(a -x)

= (a - x) / (√( a^2 - x^2)) = a/√(a^2 - x^2) - x/√( a^2 - x^2)

Putt x = a cos (A) , dx = - sinAdA

for the limits { x = a => A = π } and { x=a => A = 0 }


A/Q I= ∫ [a, -a ] √ { (a-x) / (a+x) } dx

= ∫[ a, -a] {a/√(a^2 - x^2) } dx - ∫ [a ,-a ] {x/√( a^2 - x^2)} dx

= ∫[ 0, π]{ a/√a(1 - cos^2A) }(- asinA) dA - ∫[ 0, π]{ acosA/√a(1 - cos^2A) }(- asinA) dA

= ∫[ 0, π]{ a^2sinA/asinA} dA - ∫[ 0, π]{ a^2sinAcosA/asinA} dA


=a * ∫[ 0, π] dA -a ∫[ 0, π] cosAdA

= [aA][ 0, π] - [ asinA] [ 0, π]

= a* π