Question

If $y=\int \frac{dx}{(1+x^2{)}^{\frac{3}{2}}}$ and $y=0$ when $x=0$ , the value of $y$ when $x=1$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\sqrt{2}$
• C. $2\sqrt{2}$
• D. none of these

Answer 1

Answer: (A)

$\frac{1}{\sqrt{2}}$

Let $x=\tan \theta .$ Then $dx={\sec }^2\theta d\theta$. Now,
$y=\int \frac{dx}{(1+x^2{)}^{\frac{3}{2}}}=\int \frac{{\sec }^2\theta }{(1+{\tan }^2\theta {)}^{\frac{3}{2}}}d\theta$
$=\int \frac{{\sec }^2\theta }{({\sec }^2\theta {)}^{\frac{3}{2}}}d\theta$
$=\int \frac{{\sec }^2\theta }{\left({\sec }^3\theta \right)}d\theta =\int \frac{d\theta }{\sec \theta }=\int \cos \theta d\theta$
Hence, $y=\sin \theta +c=\frac{x}{\sqrt{1+x^2}}+c$ (1)
$[\because \tan \theta =x=\frac{x}{1},\sin \theta =\frac{x}{\sqrt{1^2+x^2}}]$
Given when $x=0,y=0$ from equation (1),
$0=0+c$ or $c=0$
Thus, from equation (1), $y=\frac{x}{\sqrt{1+x^2}}$
Hence, when $x=1,y=\frac{1}{\sqrt{2}}$