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STD 12 - 7.2 definite integral — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 7.2 definite integral4 Marks
MCQ
$\int_0^1 {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}} \,dx = $
  • A
    $\frac{{{\pi ^2}}}{8}$
  • B
    $\frac{{{\pi ^2}}}{{16}}$
  • C
    $\frac{{{\pi ^2}}}{4}$
  • ✓
    $\frac{{{\pi ^2}}}{{32}}$
✓

Answer

Correct option: D.
$\frac{{{\pi ^2}}}{{32}}$
d
(d) Put $t = {\tan ^{ - 1}}x$

$\Rightarrow dt = \frac{1}{{1 + {x^2}}}dx,$ then

$\int_0^1 {\frac{{{{\tan }^{ - 1}}x}}{{1 + {x^2}}}dx = \int_0^{\pi /4} {t\,dt = \left[ {\frac{{{t^2}}}{2}} \right]_0^{\pi /4} = \frac{{{\pi ^2}}}{{32}}} } $.

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