Question

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

• A. $\sqrt{\frac{2}{3}}$
• B. $\sqrt{2}$
• C. $\frac{1}{2}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{1}{\sqrt{2}}$

Let, $x=\tan \theta$
$dx={\sec }^2\theta d\theta$
$\int \frac{dx}{(1+x^2{)}^{\frac{3}{2}}}=\int \frac{{\sec }^2\theta d\theta }{({\sec }^2\theta {)}^{\frac{3}{2}}}$
$=\int \cos \theta d\theta =\sin \theta +c=\frac{x}{\sqrt{1+x^2}}+c$
$(x,y)=(0,0)$
$\Rightarrow c=0$
When $y=1,$
$=\frac{x}{\sqrt{1+x^2}}$
$=\frac{1}{\sqrt{2}}$