Question

${\tan }^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)$ can be integrated by substituting

• A. $x=\tan \theta$
• B. $x=\cos \theta$
• C. $x=\sin \theta$
• D. $x=\sec \theta$

Answer 1

The correct option is A $x=\tan \theta$

Substitute, $x=\tan \theta \Rightarrow dx={\sec }^2\theta d\theta$
$\int {\tan }^{-1}\left(\frac{3\tan \theta -(\tan \theta {)}^3}{1-3(\tan \theta {)}^2}\right){\sec }^2\theta d\theta$

$\Rightarrow \int {\tan }^{-1}\left(\tan 3\theta \right){\sec }^2\theta d\theta =\int 3\theta {\sec }^2\theta d\theta$

Now, it can be integrated using integration by parts.
Hence, option 'A' is correct.