Question

If

$\alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} dx$,

then the value of $\sqrt{7} \tan (\frac{2\alpha \sqrt{7}}{\pi})$ is ______.

(Here, the inverse trigonometric function $\tan^{-1} x$ assumes values in $(-\frac{\pi}{2}, \frac{\pi}{2})$.)

Answer 1

21

Answer 2

The integral is given by $\alpha = \int_{\frac{1}{2}}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} dx$. Let the denominator be $D(x) = 2x^2 - 3x + 2$. The limits of integration are $1/2$ and $2$. Notice that $1/2 \times 2 = 1$. This suggests a substitution $x = 1/t$. Let $x = 1/t$, then $dx = -1/t^2 dt$. When $x = 1/2$, $t = 2$. When $x = 2$, $t = 1/2$. Substituting these into the integral: $\alpha = \int_{2}^{1/2} \frac{\tan^{-1} (1/t)}{2(1/t)^2 - 3(1/t) + 2} (-\frac{1}{t^2}) dt$ $\alpha = \int_{2}^{1/2} \frac{\tan^{-1} (1/t)}{(2 - 3t + 2t^2)/t^2} (-\frac{1}{t^2}) dt$ $\alpha = \int_{2}^{1/2} \frac{\tan^{-1} (1/t)}{2t^2 - 3t + 2} (-1) dt$ $\alpha = \int_{1/2}^{2} \frac{\tan^{-1} (1/t)}{2t^2 - 3t + 2} dt$. Replacing the dummy variable $t$ with $x$, we get: $\alpha = \int_{1/2}^{2} \frac{\tan^{-1} (1/x)}{2x^2 - 3x + 2} dx$.

Now we have two expressions for $\alpha$: (1) $\alpha = \int_{1/2}^{2} \frac{\tan^{-1} x}{2x^2 - 3x + 2} dx$ (2) $\alpha = \int_{1/2}^{2} \frac{\tan^{-1} (1/x)}{2x^2 - 3x + 2} dx$

Adding (1) and (2): $2\alpha = \int_{1/2}^{2} \frac{\tan^{-1} x + \tan^{-1} (1/x)}{2x^2 - 3x + 2} dx$. For $x > 0$, we have $\tan^{-1} x + \tan^{-1} (1/x) = \pi/2$. Since the integration interval is $[1/2, 2]$, $x$ is always positive. $2\alpha = \int_{1/2}^{2} \frac{\pi/2}{2x^2 - 3x + 2} dx$ $2\alpha = \frac{\pi}{2} \int_{1/2}^{2} \frac{1}{2x^2 - 3x + 2} dx$ $\alpha = \frac{\pi}{4} \int_{1/2}^{2} \frac{1}{2x^2 - 3x + 2} dx$.

Now we evaluate the integral $I = \int_{1/2}^{2} \frac{1}{2x^2 - 3x + 2} dx$. Complete the square in the denominator: $2x^2 - 3x + 2 = 2(x^2 - \frac{3}{2}x + 1) = 2[(x - \frac{3}{4})^2 - (\frac{3}{4})^2 + 1] = 2[(x - \frac{3}{4})^2 - \frac{9}{16} + 1] = 2[(x - \frac{3}{4})^2 + \frac{7}{16}]$. So, $I = \int_{1/2}^{2} \frac{1}{2[(x - \frac{3}{4})^2 + \frac{7}{16}]} dx = \frac{1}{2} \int_{1/2}^{2} \frac{1}{(x - \frac{3}{4})^2 + (\frac{\sqrt{7}}{4})^2} dx$. This integral is of the form $\int \frac{1}{u^2 + a^2} du = \frac{1}{a} \tan^{-1} (\frac{u}{a})$. Here $u = x - \frac{3}{4}$ and $a = \frac{\sqrt{7}}{4}$. $I = \frac{1}{2} \left[ \frac{1}{\frac{\sqrt{7}}{4}} \tan^{-1} \left( \frac{x - \frac{3}{4}}{\frac{\sqrt{7}}{4}} \right) \right]_{1/2}^{2} = \frac{1}{2} \left[ \frac{4}{\sqrt{7}} \tan^{-1} \left( \frac{4x - 3}{\sqrt{7}} \right) \right]_{1/2}^{2} = \frac{2}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{4x - 3}{\sqrt{7}} \right) \right]_{1/2}^{2}$.

Evaluate at the limits: $I = \frac{2}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{4(2) - 3}{\sqrt{7}} \right) - \tan^{-1} \left( \frac{4(1/2) - 3}{\sqrt{7}} \right) \right]$ $I = \frac{2}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{5}{\sqrt{7}} \right) - \tan^{-1} \left( \frac{-1}{\sqrt{7}} \right) \right]$. Using $\tan^{-1}(-y) = -\tan^{-1}(y)$: $I = \frac{2}{\sqrt{7}} \left[ \tan^{-1} \left( \frac{5}{\sqrt{7}} \right) + \tan^{-1} \left( \frac{1}{\sqrt{7}} \right) \right]$. Using $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left( \frac{A + B}{1 - AB} \right)$ for $AB < 1$: $A = \frac{5}{\sqrt{7}}$, $B = \frac{1}{\sqrt{7}}$. $AB = \frac{5}{7} < 1$. $\tan^{-1} \left( \frac{5}{\sqrt{7}} \right) + \tan^{-1} \left( \frac{1}{\sqrt{7}} \right) = \tan^{-1} \left( \frac{5/\sqrt{7} + 1/\sqrt{7}}{1 - (5/\sqrt{7})(1/\sqrt{7})} \right) = \tan^{-1} \left( \frac{6/\sqrt{7}}{1 - 5/7} \right) = \tan^{-1} \left( \frac{6}{\sqrt{7}} \times \frac{7}{2} \right) = \tan^{-1} (3\sqrt{7})$. So, $I = \frac{2}{\sqrt{7}} \tan^{-1} (3\sqrt{7})$.

Now substitute $I$ back into the expression for $\alpha$: $\alpha = \frac{\pi}{4} I = \frac{\pi}{4} \times \frac{2}{\sqrt{7}} \tan^{-1} (3\sqrt{7}) = \frac{\pi}{2\sqrt{7}} \tan^{-1} (3\sqrt{7})$.

We need to find the value of $\sqrt{7} \tan (\frac{2\alpha \sqrt{7}}{\pi})$. Let's calculate the argument of the tangent function: $\frac{2\alpha \sqrt{7}}{\pi} = \frac{2 \times (\frac{\pi}{2\sqrt{7}} \tan^{-1} (3\sqrt{7})) \times \sqrt{7}}{\pi} = \frac{\pi \tan^{-1} (3\sqrt{7})}{\pi} = \tan^{-1} (3\sqrt{7})$.

The expression we need to evaluate is $\sqrt{7} \tan (\tan^{-1} (3\sqrt{7}))$. Since the range of $\tan^{-1} x$ is $(-\pi/2, \pi/2)$, and $3\sqrt{7}$ is a real number, $\tan(\tan^{-1}(3\sqrt{7})) = 3\sqrt{7}$. The value is $\sqrt{7} \times (3\sqrt{7}) = 3 \times (\sqrt{7})^2 = 3 \times 7 = 21$.

The final answer is 21.