$\int(\sqrt{1+\sin (2 x)}) d x=$

$\int(\sqrt{1+\sin (2 x)}) d x=$
  1. $\cos (x)+\sin (x)+c$
  2. $\cos (x)-\sin (x)+c$
  3. $\sin (x)-\cos (x)+c$
  4. Can be option b or c depending on value of x

Solution

$ \begin{aligned} & (b, c) I=\int \sqrt{1+\sin 2 x} d x=\int|\sin x+\cos x| d x \\ & = \begin{cases}\int(\sin x+\cos x) d x, & \text { if } \sin x+\cos x \geq 0 \\ -\int(\sin x+\cos x) d x, & \text { if } \sin x+\cos x < 0\end{cases} \\ & \therefore \quad \int \begin{array}{ll} |\sin x+\cos x| d x \end{array} \\ & = \begin{cases}\sin x-\cos x+C, & \text { if } \sin x+\cos x \geq 0 \\ \cos x-\sin x+C, & \text { if } \sin x+\cos x < 0\end{cases} \end{aligned} $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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