Question

$y=\int \cos \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}dx$ is an equation of a family of

• A. straight lines
• B. circles
• C. ellipses
• D. parabolas

Answer 1

The correct option is D parabolas

$y=\int \cos \left\{2{\tan }^{-1}\sqrt{\frac{1-x}{1+x}}\right\}dx$
Taking $u={\tan }^{-1}\sqrt{\frac{1-x}{1+x}}$
$\tan u=\sqrt{\frac{1-x}{1+x}}$
$(1+x){\tan }^{2}u=1-x$
$x=\frac{1-{\tan }^{2}u}{1+{\tan }^{2}u}=\cos 2u$
$dx=-2\sin 2udu$

$\therefore y=\int \cos 2u(-2\sin 2u)du$
$=-\int \sin 4udu$
$=\frac{\cos 4u}{4}+c$
$\cos 4u=2{\cos }^{2}2u-1=2x^2-1$

$\Rightarrow y=\frac{x^2}{2}+c$ which represents the family of parabolas.