If the value of the integral _0^1/2x^2(1-x^2)^3/2,dx is k6, then k is equal to:

Question

If the value of the integral $\displaystyle\int_0^{1/2}\frac{x^2}{(1-x^2)^{3/2}}\,dx$ is $\dfrac k6$, then $k$ is equal to:

Accepted Answer

Let $x=\sin\theta\Rightarrow dx=\cos\theta\,d\theta$; when $x=\dfrac12$, $\theta=\dfrac\pi6$:
$$\int_0^{1/2}\frac{x^2}{(1-x^2)^{3/2}}dx=\int_0^{\pi/6}\frac{\sin^2\theta\cos\theta}{\cos^3\theta}d\theta=\int_0^{\pi/6}\tan^2\theta\,d\theta=\int_0^{\pi/6}(\sec^2\theta-1)\,d\theta$$

$$=\left[\tan\theta-\theta\right]_0^{\pi/6}=\frac1{\sqrt3}-\frac\pi6=\frac{2\sqrt3-\pi}6$$

$$\therefore\frac k6=\frac{2\sqrt3-\pi}6\Rightarrow k=2\sqrt3-\pi$$

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