Q118 / 160 · TS EAMCET 2024 (09 May Shift 2) · Mathematics

$\int_0^{32 \pi} \sqrt{1-\cos 4 x} d x=$

  • A
    $16 \sqrt{2}$
  • B
    $32 \sqrt{2}$
  • C
    $128 \sqrt{2}$
  • Correct answer
    D
    $64 \sqrt{2}$

Solution

$\int_0^{32 \pi} \sqrt{1-\cos 4 x} d x$ $\begin{aligned} & \int_0^{32 \pi}|\sqrt{2} \sin 2 x| d x=\sqrt{2} \int_0^{32 \pi}|\sin 2 x| d x \\ & =64 \times \sqrt{2} \int_0^{\pi / 2}(\sin 2 x) d x \\ & =\frac{64 \sqrt{2}}{2}[(-\cos 2 x)]_0^{\pi / 2}=64 \sqrt{2}\end{aligned}$

Hence, the correct answer is option: 'D'.

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