The given function is \(\displaystyle \sqrt{1 – \sin 2x} \) and we need to compute the integral of this function with respect to \( x \), i.e., compute \(\displaystyle \int \sqrt{1 – \sin 2x} \, dx \).
Solution:
First, simplify the Integrand:
\( \begin{aligned} &1 – \sin 2x \\ &= \sin^2 x + \cos^2 x \, – 2 \sin x \cos x, \quad \because \sin^2 x + \cos^2 x = 1 \text{ and } \sin 2x = 2 \sin x \cos x \\ &= (\sin x \,- \cos x)^2 \\ \end{aligned} \)Thus, \( \sqrt{1 \,- \sin 2x} = \sqrt{(\sin x \, – \cos x)^2} = |\sin x \,- \cos x| \).
When \(\sin x \geq \cos x, |\sin x – \cos x| = \sin x – \cos x \). Thus,
\( \begin{aligned} &\int (\sin x \,- \cos x) \, dx \\ &= -\cos x \,- \sin x + C, \quad \text{C is a constant} \\ &= -\left(\cos x + \sin x \right) + C \end{aligned} \)When \(\sin x < \cos x, |\sin x – \cos x| = \cos x – \sin x\). Thus,
\( \begin{aligned} &\int (\cos x \,- \sin x) \, dx \\ &= \sin x \,+ \cos x + C, \quad \text{C is a constant} \\ &= \left(\cos x + \sin x \right) + C \end{aligned} \)Therefore, \( \int \sqrt{1 – \sin 2x} \, dx = -\text{sgn}(\sin x – \cos x) (\cos x + \sin x) + C \). where \(\text{sgn} \) is the sign function.
Thus, the integral of the given function is
\( \boxed{\int \sqrt{1 \,- \sin 2x} \, dx = -\text{sgn}(\sin x \,- \cos x) (\cos x + \sin x) + C } \)Please let me know in the comments if you find any errors in this solution.