A) \[2(\sqrt{2}+1)\]
B) \[\sqrt{2}-1\]
C) \[2(\sqrt{2}-1)\]
D) 0
Correct Answer: C
Solution :
Let \[I=\int_{0}^{\pi /2}{|\sin x-\cos x|}dx\] \[=\int_{0}^{\pi /4}{(\cos x-\sin x)}dx\] \[+\int_{\pi /4}^{\pi /2}{(\sin x-\cos x)dx}\] \[=[\sin x+\cos x]_{0}^{\pi /4}+[-\cos x-\sin x]_{\pi /4}^{\pi /2}\] \[=\left[ \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-(1) \right]+\left[ 0-1\left( -\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}} \right) \right]\] \[=[\sqrt{2}-1-1+\sqrt{2}]=2(\sqrt{2}-1)\] Alternate Solution: Let \[I=\int_{0}^{\pi /2}{|\sin x-\cos x|dx}\] \[=\sqrt{2}\int_{0}^{\pi /2}{\left| \cos \frac{\pi }{4}\sin x-\sin \frac{\pi }{4}\cos x \right|}dx\] \[=\sqrt{2}\int_{0}^{\pi /2}{\left| \sin \left( x-\frac{\pi }{4} \right) \right|}dx\] Put, \[x-\frac{\pi }{4}=t\]\[\Rightarrow \]\[dx=dt\] \[=\sqrt{2}\int_{-\pi /4}^{\pi /4}{|\sin t|}\,dt\] \[=2\sqrt{2}\int_{0}^{\pi /4}{\sin t\,dt}\] \[=2\sqrt{2}[-\cos t]_{0}^{\pi /4}\] \[=-2\sqrt{2}\left[ \frac{1}{\sqrt{2}}-1 \right]\] \[=2(\sqrt{2}-1)\]
You need to login to perform this action.
You will be redirected in
3 sec
