integral sqrt(25-4x^2) / x dx

Question

Step by step so i can understand how to get the answer.

Answer 1

 ∫ sqrt(25 - x²) dx 

Let x = 5sin(θ) 
dx = 5cos(θ) dθ 

= ∫ sqrt(25 - [5sin(θ)]²) 5cos(θ) dθ 
= 5 ∫ sqrt(25 - 25sin²(θ)) cos(θ) dθ 
= 5 ∫ sqrt[25(1 - sin²(θ))] cos(θ) dθ 
= 5 ∫ 5sqrt[(1 - sin²(θ))] cos(θ) dθ 
= 25 ∫ sqrt[(1 - sin²(θ))] cos(θ) dθ 
= 25 ∫ sqrt[cos²(θ)] cos(θ) dθ 
= 25 ∫ cos(θ)cos(θ) dθ 
= 25 ∫ cos²(θ) dθ 
= 25 ∫ [1 + cos(2θ)]/2 dθ 
= 25/2 ∫ [1 + cos(2θ)] dθ 
= 25/2 [θ + sin(2θ)/2] + C 
= 25/2 [θ + cos(θ)sin(θ)] + C 

Recall that x = 5sin(θ) 
Thus θ = arcsin(x/5) 

= 25/2 [arcsin(x/5) + cos(arcsin(x/5))sin(arcsin(x/5))] + C 
= 25/2 [arcsin(x/5) + (x/5)cos(arcsin(x/5))] + C 
= 25/2 [arcsin(x/5) + (x/5)sqrt[1 - (x/5)²]] + C 
= 25/2 [arcsin(x/5) + (x/5)sqrt(1 - x²/25)] + C