Integration in calculus

Question

Find or evaluate the integral by completing the square. (Round your answer to three decimal places.)

Boundary is from 2 to 3. 

 

∫ (2x-3)/ (√(4x-x²) dx

Answer 1

Using square completion, write x² − 4x as x² − 4x + 4 − 4 or (x −2)² −4.

Then 4x −x² equals 4 − (x −2)². Set u equal to x −2 & du = dx. Also, 2u = 2x − 4

and 2u + 1 = 2x − 3.Rewrite the integral as ∫[(2u+1)du/√(4−u²)].

Draw a right triangle with the 90° angle at bottom right; mark the angle to the left of the 90° angle as θ. Mark the side opposite θ as u, the side opposite 90° as 2, and the bottom side as √(4−u²). This will give, in agreement with the Pythagorean Theorem, [√(4−u²)]² + u² = 2² or 4 = 4.

Then, by Trigonometric Substitution, obtain

sin θ = u/2 or 0.5u & u = 2sin θ & du = 2cos θdθ.

Rewrite the integral again as ∫[(4sin θ+1)2cos θdθ/√(4−4sin²θ)] which will simplify to

∫[(4sin θcos θ+cos θ)dθ/cos θ] or ∫(4sin θ+1)dθ.

From ∫(4sin θ+1)dθ, integrate to -4cos θ+θ.

-4cos θ+θ translates back to -4cos (arcsin 0.5u) + (arcsin 0.5u) and then back to

-4cos (arcsin (0.5x−1))+(arcsin (0.5x−1)).

[-4cos (arcsin (0.5x−1))+(arcsin (0.5x−1))|_{(from x=2 to x=3)}], using Radians as the unit of angle measure,

will give (-4√3/2+π/6)−(-4cos 0+0) or -2√3+π/6+4 equivalent to 1.059 square units.