The value of the definite integral ₀^ 1 5 + 4 x , dx is equal to ...

Step-by-step solution

Let I = ₀^ 1 5 + 4 x , dx Using the half-angle substitution x = 1 - ^2(x/2) 1 + ^2(x/2) and dx = 2 , dt 1 + t^2 where t = (x/2) The limits change from x = 0 to t = 0 I = ₀^ 1 5 + 4 ( 1 - t^2 1 + t^2 ) 2 , dt 1 + t^2 I = ₀^ 2 , dt 5(1 + t^2) + 4(1 - t^2) I = ₀^ 2 , dt 9 + t^2 I = 2 [ 1 3 ⁻¹ ( t 3 ) ]₀^ I = 2 3 ( 2 - 0 ) = 3 Answer: 3