Mathematics

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

$2\surd{2}$

##### 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
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 84

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