Question

$\underset{0}{\overset{1/2}{∫}}\frac{dx}{(1+x^2)\sqrt{1−x^2}}$ is equal to

• A. $\frac{1}{\sqrt{2}}{\tan }^{−1}\sqrt{\frac{2}{3}}$
• B. $\frac{2}{\sqrt{2}}{\tan }^{−1}\left(\frac{3}{\sqrt{2}}\right)$
• C. $\frac{\sqrt{2}}{2}{\tan }^{−1}\left(\frac{3}{2}\right)$
• D. $\frac{\sqrt{2}}{2}{\tan }^{−1}\left(\frac{\sqrt{3}}{2}\right)$

Answer 1

Answer: (A)

Put $x=\sin θ$
$⇒cosθdθ=dx$
When $x=0,θ=0$ and when $x=\frac{1}{2},θ=\frac{π}{6}$
$∴I=\underset{0}{\overset{\frac{π}{6}}{∫}}\frac{\cos θdθ}{(1+{\sin }^2θ)\cos θ}$
$=\underset{0}{\overset{\frac{π}{6}}{∫}}\frac{{\sec }^2θ}{{\sec }^2θ+{\tan }^2θ}dθ$
$=\underset{0}{\overset{\frac{π}{6}}{∫}}\frac{{\sec }^2θ}{1+(\sqrt{2}\tan θ{)}^2}dθ$
$=\frac{1}{\sqrt{2}}\underset{0}{\overset{\frac{π}{6}}{∫}}\frac{\sqrt{2}{\sec }^2θ}{1+(\sqrt{2}\tan θ{)}^2}dθ$
$=\frac{1}{\sqrt{2}}{\tan }^{−1}(\sqrt{2}\tan θ){â}_0^{\frac{Ï€}{6}}=\frac{1}{\sqrt{2}}{\tan }^{−1}\left(\sqrt{\frac{2}{3}}\right)$