Subject: Mathematics — Definite Integration
Question: If $\alpha = \int_{1/2}^{2}\frac{\tan^{- 1}x}{2x^{2} - 3x + 2}\, dx,$ then the value of $\sqrt{7}\tan\left( \frac{2\alpha\sqrt{7}}{\pi} \right)$ is \_\_\_\_\_\_\_\_\_\_.
Here, the inverse trigonometric function $\tan^{- 1}x$ assumes values
in $\left( - \frac{\pi}{2},\frac{\pi}{2} \right)$.
Answer: $21$
Solution: $\alpha = \int_{1/2}^{2}\frac{\tan^{- 1}x}{2x^{2} - 3x + 2}\, dx.$
Put
$$z = \frac{1}{x}.$$
$$\alpha = \int_{2}^{1/2}\frac{\cot^{- 1}z\,( - dz)}{2z^{2} - 3z + 2}.$$
$$\alpha = \int_{1/2}^{2}\frac{\frac{\pi}{2} - \tan^{- 1}z}{2z^{2} - 3z + 2}\, dz.$$
$$2\alpha = \frac{\pi}{2}\int_{1/2}^{2}\frac{1}{2z^{2} - 3z + 2}\, dz.$$
$$2\alpha = \frac{\pi}{2} \times \frac{1}{2}\int_{1/2}^{2}\frac{1\, dz}{\left( z - \frac{3}{4} \right)^{2} + \left( \frac{\sqrt{7}}{4} \right)^{2}}.$$
Now using
$$\int\frac{1}{x^{2} + a^{2}}\, dx = \frac{1}{a}\tan^{- 1}\left( \frac{x}{a} \right) + C.$$
We have
$$2\alpha = \frac{\pi}{\sqrt{7}}\left\lbrack \tan^{- 1}\left( \frac{5}{\sqrt{7}} \right) + \tan^{- 1}\left( \frac{1}{\sqrt{7}} \right) \right\rbrack.$$
$$\Rightarrow \frac{2\alpha\sqrt{7}}{\pi} = \tan^{- 1}\left( 3\sqrt{7} \right).$$
$$\tan\left( \frac{2\alpha\sqrt{7}}{\pi} \right) = 3\sqrt{7}.$$
$$\text{Now,}\quad\quad\sqrt{7}\tan\left( \frac{2\alpha\sqrt{7}}{\pi} \right) = 21.$$