MCQ
$\int_{}^{} {\frac{{\sin 2x}}{{{a^2} + {b^2}{{\sin }^2}x}}} \;dx = $
  • $\frac{1}{{{b^2}}}\log ({a^2} + {b^2}{\sin ^2}x) + c$
  • B
    $\frac{1}{b}\log ({a^2} + {b^2}{\sin ^2}x) + c$
  • C
    $\log ({a^2} + {b^2}{\sin ^2}x) + c$
  • D
    ${b^2}\log ({a^2} + {b^2}{\sin ^2}x) + c$

Answer

Correct option: A.
$\frac{1}{{{b^2}}}\log ({a^2} + {b^2}{\sin ^2}x) + c$
a
(a) Put ${a^2} + {b^2}{\sin ^2}x = t \Rightarrow {b^2}\sin 2x\,dx = dt,$ then
$\int_{}^{} {\frac{{\sin 2x}}{{{a^2} + {b^2}{{\sin }^2}x}}\,dx = \frac{1}{{{b^2}}}\int_{}^{} {\frac{{dt}}{t} = \frac{1}{{{b^2}}}\log t + c} } $
$ = \frac{1}{{{b^2}}}\log ({a^2} + {b^2}{\sin ^2}x) + c.$

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