Definite Integration Question 81
Question: $ \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 $ is equal to
[RPET 2000]
Options:
A) $ \sqrt{2}-2 $
B) $ 2\sqrt{2}-2 $
C) $ 3\sqrt{2}-2 $
D) $ 4\sqrt{2}-2 $
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Answer:
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=[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 ]-[ -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}-( \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} ) ]+[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1 ] $
$ 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 $ .
BETA
