Explanation

(b)
Let $I = \int_{0}^{1}{\frac{x^{7}}{\sqrt{1 - x^{4}}}dx =}\int_{0}^{1}\frac{x^{6}x\ dx}{\sqrt{1 - x^{4}}}$
Put $x^{2} = \sin{\theta\ \ \Rightarrow \ 2x\ dx = \cos{\theta\ d\theta}}$
$\therefore\ I = \frac{1}{2}\int_{0}^{\pi/2}\frac{\sin^{3}{\theta\cos{\theta\ d\theta}}}{\sqrt{1 - \sin^{2}\theta}}$
$\Rightarrow \ I = \frac{1}{2}\int_{0}^{\pi/2}{\frac{\sin^{3}{\theta\ cos\theta\ d\theta}}{\cos\theta} = \frac{1}{2}\int_{0}^{\pi/2}{\sin^{3}{\theta\ d\theta}}}$
$\Rightarrow \ I = \frac{1}{2}\frac{\sqrt{2}\sqrt{\left( \frac{1}{2} \right)}}{\sqrt{2}.\sqrt{\left( \frac{5}{2} \right)}} = \frac{\sqrt{\left( \frac{1}{2} \right)}}{4 \bullet \frac{3}{2} \bullet \frac{1}{2} \bullet \sqrt{\left( \frac{1}{2} \right)}} = \frac{1}{3}$