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