int_0^{32 pi} sqrt{1-cos 4 x} , dx = (in sqrt | vedclass

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(D) We know that $1 - \cos 4x = 2 \sin^2(2x)$.Therefore, $\sqrt{1 - \cos 4x} = \sqrt{2 \sin^2(2x)} = \sqrt{2} |\sin 2x|$.The integral becomes $\int_0^

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$64$

Vedclass · Answer

(D) We know that $1 - \cos 4x = 2 \sin^2(2x)$.Therefore, $\sqrt{1 - \cos 4x} = \sqrt{2 \sin^2(2x)} = \sqrt{2} |\sin 2x|$.The integral becomes $\int_0^{32 \pi} \sqrt{2} |\sin 2x| \, dx = \sqrt{2} \int_0^{32 \pi} |\sin 2x| \, dx$.Since $|\sin 2x|$ is a periodic function with period $\frac{\pi}{2}$, and the interval $[0, 32 \pi]$ contains $\frac{32 \pi}{\pi/2} = 64$ periods.Thus, $\int_0^{32 \pi} |\sin 2x| \, dx = 64 \int_0^{\pi/2} \sin 2x \, dx$.Evaluating the integral: $64 \left[ -\frac{\cos 2x}{2} \right]_0^{\pi/2} = 64 \left( -\frac{\cos \pi}{2} - (-\frac{\cos 0}{2}) \right) = 64 \left( \frac{1}{2} + \frac{1}{2} \right) = 64$.Multiplying by the constant $\sqrt{2}$, the final result is $64 \sqrt{2}$.

$\int_0^{32 \pi} \sqrt{1-\cos 4 x} \, dx =$ (in $\sqrt{2}$)

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