Need complete the square inside the square root the denominator xnow

Solved Step by Step With Explanation-Trig Sub Integral Solution

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Solved Step by Step With Explanation-Trig Sub Integral Solution

√(4x - x^2) = √(4 - x^2) * √x

Now, we can rewrite the integral as:

Now, we'll find dx in terms of dθ:

dx = 2cos(θ) dθ

∫(2x - 3)/(2cos(θ) * √x) * 2cos(θ) dθ

Step 3: Simplify the integral

∫(2x - 3)/(√x) dθ = ∫(2(2sin(θ)) - 3)/(√(2sin(θ))) * 2cos(θ) dθ

= ∫(4sin(θ) - 3)/(√(2sin(θ))) * 2cos(θ) dθ

Let's evaluate each integral separately:

a. ∫(4sin(θ))/(√(2sin(θ))) cos(θ) dθ:

Now, this is a simpler integral:

= 8∫(u)/(√2√u) du

= (4√2/3)(sin^(3/2)(θ/2)) + C

b. ∫(3)/(√(2sin(θ))) cos(θ) dθ:

= 3∫t^(-1) dt

= 3ln|t| + C

∫(2x - 3)/√(4x - x^2) dx = 2(4√2/3)(sin^(3/2)(θ/2)) - 3ln|√(2sin(θ))| + C

Step 7: Convert back to the original variable x

∫(2x - 3)/√(4x - x^2) dx = 2(4√2/3)(sin^(3/2)(arcsin(x/2)/2)) - 3ln|√(2sin(arcsin(x/2)))| + C

= (8√2/3)(x/2)^(3/2) - 3ln|√(2(x/2))| + C