Question:
\(\int_{0}^{\pi} \sqrt{\frac{1+\cos 2x}{2}}\, \mathrm{d}x\) is ________.
\(\int_{0}^{\pi} \sqrt{\frac{1+\cos 2x}{2}}\, \mathrm{d}x\) is ________.
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Use \(1+\cos2x=2\cos^2x\), so the integrand becomes \(|\cos x|\).
Updated On: Nov 12, 2026
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Correct Answer: 2
Solution and Explanation
\(\frac{1+\cos2x}{2}=\cos^2x\), so the integrand is \(\sqrt{\cos^2x}=|\cos x|\).
\(\int_0^{\pi}|\cos x|dx=\int_0^{\pi/2}\cos x\,dx+\int_{\pi/2}^{\pi}(-\cos x)\,dx\).
\(=[\sin x]_0^{\pi/2}-[\sin x]_{\pi/2}^{\pi}=(1-0)-(0-1)=2\).
Answer: 2
\(\int_0^{\pi}|\cos x|dx=\int_0^{\pi/2}\cos x\,dx+\int_{\pi/2}^{\pi}(-\cos x)\,dx\).
\(=[\sin x]_0^{\pi/2}-[\sin x]_{\pi/2}^{\pi}=(1-0)-(0-1)=2\).
Answer: 2
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