MCQ
$\int_0^{\pi / 2}|\sin x-\cos x| d x=$
- A$0$
- ✓$2(\sqrt{2}-1)$
- C$\sqrt{2}-1$
- D$2(\sqrt{2}+1)$
✓
Answer
Correct option: B.
$2(\sqrt{2}-1)$
(B)
$\int_0^{\pi / 2}|\sin x-\cos x| d x$
$\begin{array}{l}=-\int_0^{\pi / 4}(\sin x-\cos x) d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x \\ =-[-\cos x-\sin x]_0^{\pi / 4}+[-\cos x-\sin x]_{\pi / 4}^{\pi / 2} \\ =2(\sqrt{2}-1)\end{array}$
$\int_0^{\pi / 2}|\sin x-\cos x| d x$
$\begin{array}{l}=-\int_0^{\pi / 4}(\sin x-\cos x) d x+\int_{\pi / 4}^{\pi / 2}(\sin x-\cos x) d x \\ =-[-\cos x-\sin x]_0^{\pi / 4}+[-\cos x-\sin x]_{\pi / 4}^{\pi / 2} \\ =2(\sqrt{2}-1)\end{array}$
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