Question

\[\int\limits_0^{1/\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) dx\]

Answer 1

\[\int_0^\frac{1}{\sqrt{3}} \tan^{- 1} \left( \frac{3x - x^3}{1 - 3 x^2} \right) d x\]

\[Let x = \tan\theta,\text{ then }dx = \sec^2 \theta d\theta\]

\[\text{When, }x \to 0 ; \theta \to 0\]

\[\text{And }x \to \frac{1}{\sqrt{3}} ; \theta \to \frac{\pi}{6}\]

Therefore the integral becomes

\[ \int_0^\frac{\pi}{6} \tan^{- 1} \left( \frac{3\tan\theta - \tan^3 \theta}{1 - 3 \tan^2 \theta} \right)se c^2 \theta d\theta\]

\[ = \int_0^\frac{\pi}{6} \tan^{- 1} \left( \tan3\theta \right)se c^2 \theta d\theta\]

\[ = 3 \int_0^\frac{\pi}{6} \theta se c^2 \theta d\theta\]

\[ = 3 \left[ \theta \tan\theta \right]_0^\frac{\pi}{6} - 3 \int_0^\frac{\pi}{6} \tan\theta d\theta\]

\[ = 3 \left[ \theta \tan\theta \right]_0^\frac{\pi}{6} - 3 \left[ - \log\left( \cos\theta \right) \right]_0^\frac{\pi}{6} \]

\[\]

\[ = 3\left( \frac{\pi}{6} \times \frac{1}{\sqrt{3}} - 0 \right) + 3\left[ \log\frac{\sqrt{3}}{2} \right]\]

\[ = \frac{\pi}{2\sqrt{3}} + 3\log\frac{\sqrt{3}}{2}\]

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