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7.2 definite integral — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMaths7.2 definite integral1 Mark
MCQ
$\int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,dx = } $
  • ✓
    $\frac{1}{{20}}\log 3$
  • B
    $\log 3$
  • C
    $\frac{1}{{20}}\log 5$
  • D
    None of these
✓

Answer

Correct option: A.
$\frac{1}{{20}}\log 3$
a
(a) Let $I = \int_0^{\pi /4} {\frac{{\sin x + \cos x}}{{9 + 16\sin 2x}}\,} dx$

Put $\sin x - \cos x = t$, then 

$(\sin x + \cos x)dx = dt$

$I = \int_{ - 1}^0 {\frac{{dt}}{{9 + 16(1 - {t^2})}}} $

$= \int_{ - 1}^0 {\frac{{dt}}{{25 - 16{t^2}}}} $

$ = \frac{1}{{10}}\int_{ - 1}^0 {\left( {\frac{1}{{5 - 4t}} + \frac{1}{{5 + 4t}}} \right)dt} $

$ = \left| {\frac{1}{{10}}.\frac{1}{4}[\log (5 + 4t) - \log (5 - 4t)]\,} \right|_{ - 1}^0$

$ = \frac{1}{{40}}(\log 9 - \log 1) = \frac{1}{{20}}\log 3$.

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