Question

Evaluate: ${\int }_0^1$ tan ${}^{-1}\left(\frac{3x-x^3}{1-3x^2}\right)dx$

• A. $\frac{3\pi }{2}-3\log 2$
• B. $\frac{3\pi }{4}-\frac{3}{2}\log 2$
• C. $\frac{7\pi }{2}+3\log 2$
• D. $\frac{3\pi }{4}+\frac{3}{2}|og2$

Answer 1

The correct option is B $\frac{3\pi }{4}-\frac{3}{2}\log 2$

Let $x=\tan \theta$ $⟹dx={\sec }^2\theta d\theta$

At , $x=0,\theta =0$ and at $x=1,\theta =\frac{\pi }{4}$

And, $\frac{3x-x^3}{1-3x^2}=\frac{3\tan \theta -{\tan }^3\theta }{1-3{\tan }^2\theta }=\tan 3\theta$

$⟹{\tan }^{-1}\left(\tan 3\theta \right)=3\theta$

Now the integration becomes:-

${\int }_0^{\pi /4}3\theta {\sec }^2\theta d\theta$

Applying integration by parts:-

$={[3\theta \tan \theta -3\int \tan \theta d\theta ]}_0^{\pi /4}$

$={[3\theta \tan \theta -3\log |\sec \theta |]}_0^{\pi /4}$

$=\frac{3\pi }{4}-3\log \sqrt{2}$

$=\frac{3\pi }{4}-\frac{3}{2}\log 2$