Key Concepts and Formulas

• Trigonometric Substitution: When the integrand contains expressions of the form $\sqrt{a^2 - x^2}$, the substitution $x = a \sin\theta$ or $x = a \cos\theta$ is often effective.
• Trigonometric Identities: Key identities such as $1 - \sin^2\theta = \cos^2\theta$ and $\tan^2\theta = \sec^2\theta - 1$ are crucial for simplifying trigonometric integrals.
• Definite Integration: When performing a substitution in a definite integral, the limits of integration must be changed to correspond to the new variable. The Fundamental Theorem of Calculus is used to evaluate the definite integral.

Step-by-Step Solution

Step 1: Identify the appropriate substitution The integral is $I = \int\limits_0^{{1 \over 2}} {{{{x^2}} \over {{{\left( {1 - {x^2}} \right)}^{{3 \over 2}}}}}} dx$. The presence of the term $(1-x^2)^{3/2}$, which involves $\sqrt{1-x^2}$, suggests a trigonometric substitution. Let $x = \sin\theta$. This substitution will simplify the term $1-x^2$ to $\cos^2\theta$.

Step 2: Calculate $dx$ and change the limits of integration

• Differentiate $x = \sin\theta$ with respect to $\theta$: $dx = \cos\theta \, d\theta$
• Change the lower limit: When $x = 0$, we have $\sin\theta = 0$. The principal value of $\theta$ is $0$.
• Change the upper limit: When $x = \frac{1}{2}$, we have $\sin\theta = \frac{1}{2}$. The principal value of $\theta$ is $\frac{\pi}{6}$. So, the new limits of integration are from $0$ to $\frac{\pi}{6}$.

Step 3: Substitute into the integral and simplify Substitute $x = \sin\theta$ and $dx = \cos\theta \, d\theta$ into the integral: $I = \int\limits_0^{{\pi \over 6}} {{{{(\sin\theta)}^2} \over {{{\left( {1 - {{\sin }^2}\theta } \right)}^{{3 \over 2}}}}}} (\cos\theta \, d\theta)$ Using the identity $1 - \sin^2\theta = \cos^2\theta$: I = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\left( {{{\cos }^2}\theta } \right)}^{{3 \over 2}}}}}}\cos \theta d\theta } Since $0 \le \theta \le \frac{\pi}{6}$, $\cos\theta > 0$, so $(\cos^2\theta)^{3/2} = (\cos\theta)^3 = \cos^3\theta$. $I = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\cos }^3}\theta }}\cos \theta d\theta }$ Cancel one $\cos\theta$ term: $I = \int\limits_0^{{\pi \over 6}} {{{{{\sin }^2}\theta } \over {{{\cos }^2}\theta }}d\theta }$

Step 4: Use trigonometric identities to simplify the integrand Rewrite the integrand using $\tan\theta = \frac{\sin\theta}{\cos\theta}$: $I = \int\limits_0^{{\pi \over 6}} {{{\tan }^2}\theta \, d\theta }$ Use the identity $\tan^2\theta = \sec^2\theta - 1$: $I = \int\limits_0^{{\pi \over 6}} {(\sec^2\theta - 1)d\theta }$

Step 5: Perform the integration Integrate term by term: $I = \left[ {\tan \theta - \theta } \right]_0^{{\pi \over 6}}$

Step 6: Evaluate the definite integral using the limits Apply the Fundamental Theorem of Calculus: $I = \left( {\tan \left( {\frac{\pi }{6}} \right) - \frac{\pi }{6}} \right) - \left( {\tan(0) - 0} \right)$ Substitute the known values $\tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}$ and $\tan(0) = 0$: $I = \left( {\frac{1}{{\sqrt 3 }} - \frac{\pi }{6}} \right) - (0 - 0)$ $I = \frac{1}{{\sqrt 3 }} - \frac{\pi }{6}$ To get a common denominator, rationalize $\frac{1}{\sqrt{3}}$: $\frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$ So, $I = \frac{\sqrt{3}}{3} - \frac{\pi }{6}$ Express with a common denominator of 6: $I = \frac{2\sqrt{3}}{6} - \frac{\pi }{6} = \frac{2\sqrt{3} - \pi}{6}$

Step 7: Determine the value of k The problem states that the value of the integral is $\frac{k}{6}$. We have found that $I = \frac{2\sqrt{3} - \pi}{6}$. Comparing the two expressions: $\frac{k}{6} = \frac{2\sqrt{3} - \pi}{6}$ Therefore, $k = 2\sqrt{3} - \pi$.

Common Mistakes & Tips

• Forgetting to change limits: Always change the limits of integration when performing a substitution in a definite integral.
• Sign errors with $\cos\theta$: Be mindful of the sign of trigonometric functions in different quadrants. In this case, for $0 \le \theta \le \frac{\pi}{6}$, $\cos\theta$ is positive, so $\sqrt{\cos^2\theta} = \cos\theta$.
• Algebraic simplification: Ensure careful algebraic manipulation, especially when combining terms or rationalizing denominators.

Summary The integral was evaluated using the trigonometric substitution $x = \sin\theta$. This transformed the integrand into a simpler form involving $\tan^2\theta$, which was then integrated using the identity $\tan^2\theta = \sec^2\theta - 1$. After evaluating the definite integral, the result was compared with the given form $\frac{k}{6}$ to find the value of $k$.

The final answer is \boxed{2\sqrt 3 - \pi}.