Ignited integrals 3

Using the trigonometric substitution $x = \sin \theta, dx = \cos \theta d\theta$ we transform the integral like so:

$\int \frac{x^2}{(1 - x^2)^{\frac{3}{2}}} dx \longrightarrow \int \frac{\sin^2 \theta}{\cos^3 \theta}\cos \theta d\theta = \int \tan^2 \theta d\theta$

Using the pythagorean identity $1 + \tan^2 \theta = \sec^2 \theta,$ we rewrite the resulting integral and integrate:

$\int \tan^2 \theta d\theta = \int \sec^2 \theta - 1 d\theta = \tan \theta - \theta$

Convert this result back to the variable $x$ to obtain

$\frac{x}{\sqrt{1 - x^2}} - \arcsin x \Big|_{0}^{\frac{1}{2}} = \frac{\sqrt{3}}{3} - \frac{\pi}{6} = \frac{2\sqrt{3} - \pi}{6}$

Here, A = 2, B = 3, C = 6, so our answer is 2 + 3 + 6 = $\boxed{11}.$

$P = \displaystyle \int_{0}^{\dfrac{1}{2}} \dfrac{x\cdot 2x}{2(1-x^{2})^{\dfrac{3}{2}}}dx$
Integrating by parts,
$P = \left[ \dfrac{x}{2\sqrt{1-x^{2}}\dfrac{1}{2}}\right]_{0}^{\dfrac{1}{2}} - \displaystyle \int_{0}^{\dfrac{1}{2}}\left(\dfrac{d}{dx}x\right) \dfrac{1}{2\sqrt{1-x^{2}}\dfrac{1}{2}}dx$
$\therefore P = \dfrac{1}{\sqrt{3}} - \left[\sin^{-1} x \right]_{0}^{\dfrac{1}{2}}$
$\therefore P = \dfrac{1}{\sqrt{3}} - \dfrac{\pi}{6} = \dfrac{2\sqrt{3} - \pi}{6}$
Comparing, $A = 2 , B = 3, C = 6$
$A + B + C = 2 + 3 + 6 = 11$