MCQ
$\int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{dx}}{{1\,\, + \,\,{a^2}{{\sin }^2}x}}} $ has the value :
  • $\frac{\pi }{{2\,\sqrt {1\, + \,{a^2}} }}$
  • B
    $\frac{\pi }{{\sqrt {1\, + \,{a^2}} }}$
  • C
    $\frac{{2\,\pi }}{{\sqrt {1\, + \,{a^2}} }}$
  • D
    none

Answer

Correct option: A.
$\frac{\pi }{{2\,\sqrt {1\, + \,{a^2}} }}$
a
$I = \int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{{{\sec }^2}x\,dx}}{{1 + {{\tan }^2}x + {a^2}{{\tan }^2}x}}} $ $= \int\limits_0^{\frac{\pi }{2}} {\,\,\frac{{{{\sec }^2}x\,dx}}{{(1 + {a^2}){{\tan }^2}x + 1}}} $

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