A) \[\sqrt{2}-2\]
B) \[2\sqrt{2}-2\]
C) \[3\sqrt{2}-2\]
D) \[4\sqrt{2}-2\]
Correct Answer: D
Solution :
Let \[I=\int_{0}^{\pi /4}{(\cos x-\sin x)dx}\] \[+\int_{\pi /4}^{5\pi /4}{(\sin x-\cos x)dx}\] \[+\int_{2\pi }^{\pi /4}{(\cos x-\sin x)dx}\] \[=[\sin x+\cos x]_{0}^{\pi /4}+[-\cos x-\sin x]_{\pi /4}^{5\pi /4}\] \[+[\sin x+\cos x]_{2\pi }^{\pi /4}\] \[=\left[ \sin \frac{\pi }{4}+\cos \frac{\pi }{4}-(\sin 0+\cos 0) \right]\] \[-\left[ \cos \frac{5\pi }{4}+\sin \frac{5\pi }{4}-\left( \cos \frac{\pi }{4}+\sin \frac{\pi }{4} \right) \right]\] \[+\left[ \sin \frac{\pi }{4}+\cos \frac{\pi }{4}-(\sin 2\pi +\cos 2\pi ) \right]\] \[=\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 \right]-\left[ -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-\left( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \right) \right]\] \[+\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 \right]\] \[=\left[ \frac{2}{\sqrt{2}}-1 \right]-\left[ -\frac{4}{\sqrt{2}} \right]+\left[ \frac{2}{\sqrt{2}}-1 \right]\] \[=(\sqrt{2}-1)+2\sqrt{2}+(\sqrt{2}-1)\] \[=4\sqrt{2}-2\]
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