Mathematics

The value of $$\displaystyle \overset{2\pi}{\underset{0}{\int}}max (\sin x, \cos x)dx$$ is


ANSWER

$$2\surd{2}$$


SOLUTION
$$\displaystyle\int_{0}^{2\pi}max (\sin x, \cos x)dx=\int_{0}^{\pi/4}\cos xdx+\int_{\pi/4}^{5\pi/4}\sin x dx+\int_{5\pi/4}^{2\pi}\cos xdx$$

$$=\sin x { \left. \dfrac {  }{  }  \right|  }_{ 0 }^{ \pi/4 }-\cos x{ \left. \dfrac {  }{  }  \right|  }_{ \pi/4 }^{ 5\pi/4 }+\sin x{ \left. \dfrac {  }{  }  \right|  }_{ 5\pi/4 }^{ 2\pi }$$

$$=\dfrac{1}{\sqrt{2}}-0+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}+0+\dfrac{1}{\sqrt{2}}=\dfrac{4}{\sqrt{2}}$$
On rationalising we get

$$\dfrac{4\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}=2\sqrt{2}$$
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Single Correct Medium Published on 17th 09, 2020
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