How i can solve this integral ∫ dx / akar 25 - 16x^2 ? i really need your , you.

T ∫ 1/√(1 - x²) dx = arcsin(x) 

 ∫ 1/√(1 - x²/a) dx = √a arcsin(x/√a) 

So pull a factor of 5 out of the square root to get 
1/5 ∫ 1/√(1 - 16 x²/25) dx 
= 1/4 arcsin(4 x/5)