Question

Find k if \(\int_{0}^{\frac{1}{2}}[\frac{x^2dx}{(1-x^2)^{\frac{3}{2}}}]=\frac{k}{6}\).

Answer 1

Let's solve the definite integral step by step.

1. Set up the Integral:
We are given:

∫₀^(1/2) [x² / (1 - x²)^(3/2)] dx = k/6

2. Use Trigonometric Substitution:
Let x = sin(θ). Then dx = cos(θ) dθ.

When x = 0, sin(θ) = 0, so θ = 0.
When x = 1/2, sin(θ) = 1/2, so θ = π/6.

Substitute into the integral:

∫₀^(π/6) [sin²(θ) / (1 - sin²(θ))^(3/2)] cos(θ) dθ

3. Simplify the Integral:
Using the identity 1 - sin²(θ) = cos²(θ), we get:

∫₀^(π/6) [sin²(θ) / (cos²(θ))^(3/2)] cos(θ) dθ

∫₀^(π/6) [sin²(θ) / cos³(θ)] cos(θ) dθ

∫₀^(π/6) [sin²(θ) / cos²(θ)] dθ

∫₀^(π/6) tan²(θ) dθ

4. Use Trigonometric Identity:
We know that tan²(θ) = sec²(θ) - 1.

∫₀^(π/6) (sec²(θ) - 1) dθ

5. Integrate:
[tan(θ) - θ]₀^(π/6)

6. Evaluate the Limits:
[tan(π/6) - π/6] - [tan(0) - 0]

[1/√3 - π/6] - [0 - 0]

1/√3 - π/6

7. Relate to k/6:
We are given that the integral equals k/6.

1/√3 - π/6 = k/6

Multiply by 6:

6/√3 - π = k

2√3 - π = k

Therefore, k = 2√3 - π.