Question

The value of the integral
$\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{\left(\right(x+1{)}^2(1-x{)}^6{)}^{\frac{1}{4}}}dx$
is

Answer 1

$I=\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{\left(\right(x+1{)}^2(1-x{)}^6{)}^{\frac{1}{4}}}dx=\underset{0}{\overset{\frac{1}{2}}{\int }}\frac{1+\sqrt{3}}{(1-x)\sqrt{1-x^2}}dx$
Let $x=\sin \theta \Rightarrow dx=\cos \theta d\theta$
$I=\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{(1+\sqrt{3})\cos \theta }{\cos \theta (1-\sin \theta )}d\theta =(1+\sqrt{3})\underset{0}{\overset{\frac{\pi }{6}}{\int }}\frac{(1+\sin \theta )}{{\cos }^2\theta }d\theta$
$I=(1+\sqrt{3})(\underset{0}{\overset{\frac{\pi }{6}}{\int }}{\sec }^2\theta d\theta +\underset{0}{\overset{\frac{\pi }{6}}{\int }}\sec \theta \tan \theta d\theta )$
$I=(1+\sqrt{3})(\tan \theta {|}_0^{\frac{\pi }{6}}+\sec \theta {|}_0^{\frac{\pi }{6}})$
$I=(1+\sqrt{3})(\frac{1}{\sqrt{3}}+(\frac{2}{\sqrt{3}}-1))$
$I=(\sqrt{3}+1)(\sqrt{3}-1)=((\sqrt{3}{)}^2-1^2)=2$