If \[ \alpha=\int_{\frac{1}{2}}^{2} \frac{\tan ^{-1} x}{2 x^{2}-3 x+2} d x \] 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 \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\).)

Definite Integrals — JEE Advanced · 2025 · 18 May 2025 · Paper 2 Question with Solution

α = \int_{\frac{1}{2}}^{2} \frac{tan ^{- 1} x}{2 x ^{2} - 3 x + 2} d x

then the value of

7 tan (\frac{2 α 7}{π})

is ____.
(Here, the inverse trigonometric function

tan^{- 1} x

assumes values in

(- \frac{π}{2}, \frac{π}{2})

Answer: 21

Let

I = \int_{\frac{1}{2}}^{2} \frac{t a n ^{- 1} x}{2 x ^{2} - 3 x + 2} d x

.

Substitute

x = \frac{1}{t}

d x = - \frac{1}{t ^{2}} d t

I = \int_{2}^{\frac{1}{2}} \frac{tan ^{- 1} ( 1/ t )}{2/ t ^{2} - 3/ t + 2} (- \frac{1}{t ^{2}}) d t = \int_{\frac{1}{2}}^{2} \frac{cot ^{- 1} t}{2 t ^{2} - 3 t + 2} d t

Adding the two expressions:

2 I = \int_{\frac{1}{2}}^{2} \frac{tan ^{- 1} x + cot ^{- 1} x}{2 x ^{2} - 3 x + 2} d x = \int_{\frac{1}{2}}^{2} \frac{π /2}{2 x ^{2} - 3 x + 2} d x

So:

I = \frac{π}{4} \int_{\frac{1}{2}}^{2} \frac{d x}{2 x ^{2} - 3 x + 2} = \frac{π}{8} \int_{\frac{1}{2}}^{2} \frac{d x}{x ^{2} - \frac{3 x}{2} + 1}

Complete the square:

x^{2} - \frac{3 x}{2} + 1 = (x - \frac{3}{4})^{2} + \frac{7}{16}

I = \frac{π}{8} \cdot \frac{4}{7} [tan^{- 1} (\frac{4 ( x - 3/4 )}{7})]_{\frac{1}{2}}^{2} = \frac{π}{2 7} [tan^{- 1} \frac{4 x - 3}{7}]_{\frac{1}{2}}^{2}

Evaluating:

= \frac{π}{2 7} (tan^{- 1} \frac{5}{7} - tan^{- 1} (- \frac{1}{7})) = \frac{π}{2 7} tan^{- 1} (\frac{\frac{5}{7} + \frac{1}{7}}{1 - \frac{5}{7}})

= \frac{π}{2 7} tan^{- 1} (37)

Thus

α = \frac{π}{2 7} tan^{- 1} (37)

.

Therefore:

7 tan (\frac{2 α 7}{π}) = 7 tan (tan^{- 1} (37)) = 7 \cdot 37 = 21

JEE Advanced · 2025 · 18 May 2025 · Paper 2 · Numerical answer