MCQ
$\int_{}^{} {\frac{{\sin x + \cos x}}{{\sqrt {1 + \sin 2x} }}\;dx = } $
  • A
    $\sin x + c$
  • B
    $\cos x + c$
  • $x + c$
  • D
    ${x^2} + c$

Answer

Correct option: C.
$x + c$
c
(c) $\int_{}^{} {\frac{{\sin x + \cos x}}{{\sqrt {1 + \sin 2x} }}\,dx} = \int_{}^{} {\frac{{\sin x + \cos x}}{{\sqrt {{{(\sin x + \cos x)}^2}} }}\,dx} = \int_{}^{} {dx = x + c} $.

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