Q232 / 270 · NDA 2019 (Phase 1) · Mathematics

$\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ is equal to

  • A
    0
  • Correct answer
    B
    $2(\sqrt{2}-1)$
  • C
    $2 \sqrt{2}$
  • D
    $2(\sqrt{2}+1)$

Solution

$\int_0^{\frac{\pi}{2}}|\sin x-\cos x| d x$ $\begin{aligned} & =\int_0^{\frac{\pi}{4}}|(\sin x-\cos x)| d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}|(\sin x-\cos x)| d x \\ & =\int_0^{\frac{\pi}{4}}(\cos x-\sin x) \cdot d x+\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}(\sin x-\cos x) \cdot d x \\ & =(\sin x+\cos x)_0^{\frac{\pi}{4}}+(-\cos x-\sin x)_{\frac{\pi}{2}}^{\frac{\pi}{2}} \\ & =2 \sqrt{2}-2=2(\sqrt{2}-1)\end{aligned}$

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

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