- \(\displaystyle \int_{0}^{\pi} \sqrt {1 + \cos (2x)} ~dx= \color {blue} {2\sqrt {2}}\)
- Solution: \(\sqrt {1 + \cos (2x)} = \sqrt {2 \cos ^2 (x)} = \sqrt {2} \cos (x)\)
Solution
1. For all \(x \in [0, \pi]\), \(1 + \cos(2x) = 2\cos^2(x)\), hence \(\sqrt{1 + \cos(2x)} = \sqrt{2}|\cos(x)|\).
By the cosine double-angle formula \(\cos(2x) = 2\cos^2(x) - 1\). Taking the principal square root gives \(\sqrt{2\cos^2(x)} = \sqrt{2}\sqrt{\cos^2(x)} = \sqrt{2}|\cos(x)|\).
2. Partitioning the integral over intervals of constant sign of \(\cos(x)\): \[\int_0^\pi |\cos(x)| \, dx = \int_0^{\pi/2} \cos(x) \, dx + \int_{\pi/2}^\pi (-\cos(x)) \, dx.\]
2.1. On \([0, \pi/2]\), \(\cos(x) \ge 0\), so \(|\cos(x)| = \cos(x)\).
2.2. On \([\pi/2, \pi]\), \(\cos(x) \le 0\), so \(|\cos(x)| = -\cos(x)\).
2.3. Splitting the domain of integration at \(x = \pi/2\) and using these two identities gives the stated decomposition.
3. Computing each piece:
3.1. \(\int_0^{\pi/2} \cos(x) \, dx = [\sin(x)]_0^{\pi/2} = \sin(\pi/2) - \sin(0) = 1 - 0 = 1\).
3.2. \(\int_{\pi/2}^\pi (-\cos(x)) \, dx = [-\sin(x)]_{\pi/2}^\pi = -\sin(\pi) - (-\sin(\pi/2)) = 0 + 1 = 1\).
3.3. Thus \(\int_0^\pi |\cos(x)| \, dx = 1 + 1 = 2\).
4. Conclusion: \[\int_0^\pi \sqrt{1+\cos(2x)} \, dx = \sqrt{2} \int_0^\pi |\cos(x)| \, dx = 2\sqrt{2}.\]
4.1. By 1, \(\sqrt{1+\cos(2x)} = \sqrt{2}|\cos(x)|\), so the integral equals \(\sqrt{2}\int_0^\pi|\cos(x)|\,dx\).
4.2. By 3, \(\int_0^\pi|\cos(x)|\,dx = 2\), so the value is \(\sqrt{2}\cdot 2 = 2\sqrt{2}\).