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 the integral:

\[ \int_0^{\frac{1}{2}} \frac{x^2}{(1 - x^2)^{3/2}} \, dx = \frac{k}{6} \]

2. Use Trigonometric Substitution:
Let \( x = \sin(\theta) \), so that \( dx = \cos(\theta) \, d\theta \).
When \( x = 0 \), \( \sin(\theta) = 0 \), so \( \theta = 0 \).
When \( x = \frac{1}{2} \), \( \sin(\theta) = \frac{1}{2} \), so \( \theta = \frac{\pi}{6} \).

Substituting into the integral, we get: \[ \int_0^{\frac{\pi}{6}} \frac{\sin^2(\theta)}{(1 - \sin^2(\theta))^{3/2}} \cos(\theta) \, d\theta \]

3. Simplify the Integral:
Using the identity \( 1 - \sin^2(\theta) = \cos^2(\theta) \), we get: \[ \int_0^{\frac{\pi}{6}} \frac{\sin^2(\theta)}{\cos^3(\theta)} \cos(\theta) \, d\theta \] Simplifying further: \[ \int_0^{\frac{\pi}{6}} \frac{\sin^2(\theta)}{\cos^2(\theta)} \, d\theta \] This simplifies to: \[ \int_0^{\frac{\pi}{6}} \tan^2(\theta) \, d\theta \]

4. Use Trigonometric Identity:
We know that: \[ \tan^2(\theta) = \sec^2(\theta) - 1 \] Therefore, the integral becomes: \[ \int_0^{\frac{\pi}{6}} (\sec^2(\theta) - 1) \, d\theta \]

5. Integrate:
The integral of \( \sec^2(\theta) \) is \( \tan(\theta) \), and the integral of 1 is \( \theta \). So, we have: \[ [\tan(\theta) - \theta]_0^{\frac{\pi}{6}} \]

6. Evaluate the Limits:
Evaluating at the limits: \[ \left( \tan\left(\frac{\pi}{6}\right) - \frac{\pi}{6} \right) - \left( \tan(0) - 0 \right) \] We know that: \[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}, \quad \tan(0) = 0 \] Substituting these values: \[ \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right) - (0 - 0) \] Simplifying: \[ \frac{1}{\sqrt{3}} - \frac{\pi}{6} \]

7. Relate to \( k/6 \):
We are given that the integral equals \( \frac{k}{6} \). Thus: \[ \frac{1}{\sqrt{3}} - \frac{\pi}{6} = \frac{k}{6} \] Multiply both sides by 6: \[ 6 \times \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right) = k \] Simplifying: \[ 6 \times \frac{1}{\sqrt{3}} - \pi = k \] \[ 2\sqrt{3} - \pi = k \]

Therefore, the value of \( k \) is:
\[ k = 2\sqrt{3} - \pi \]