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$\int(\sqrt{1+\sin (2 x)}) d x=$
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Verified Answer
The correct answer is:
Can be option b or c depending on value of x
$$
\begin{aligned}
& (b, c) I=\int \sqrt{1+\sin 2 x} d x=\int|\sin x+\cos x| d x \\
& = \begin{cases}\int(\sin x+\cos x) d x, & \text { if } \sin x+\cos x \geq 0 \\
-\int(\sin x+\cos x) d x, & \text { if } \sin x+\cos x < 0\end{cases} \\
& \therefore \quad \int \begin{array}{ll}
|\sin x+\cos x| d x
\end{array} \\
& = \begin{cases}\sin x-\cos x+C, & \text { if } \sin x+\cos x \geq 0 \\
\cos x-\sin x+C, & \text { if } \sin x+\cos x < 0\end{cases}
\end{aligned}
$$
\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}
$$
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