Mathematics

# The value of the integral $\displaystyle \int_{\frac{\pi}6}^{\frac{\pi}2} \left (\dfrac {1 + \sin 2x + \cos 2x}{\sin x + \cos x}\right )dx$ is equal to

$1$

##### SOLUTION
Let
$I = \displaystyle \int_{\pi / 6}^{\pi /2} \left (\dfrac {1 + \sin 2x + \cos 2x}{\sin x + \cos x}\right )dx$

$= \displaystyle \int_{\pi / 6}^{\pi /2} \left (\dfrac {1 + 2\sin x \cos x + 2\cos^{2} x - 1}{(\sin x + \cos x)}\right )dx$

$= \displaystyle \int_{\pi / 6}^{\pi /2} \dfrac {2\cos x (\sin x + \cos x)}{(\sin x + \cos x)} dx$

$= \displaystyle \int_{\pi / 6}^{\pi /2} 2\cos x dx = 2[\sin x]_{\pi / 6}^{\pi /2}$

$= 2\left (\sin \dfrac {\pi}{2} -\sin \dfrac {\pi}{6}\right ) = 2\left (1 - \dfrac {1}{2}\right ) = 2\times \dfrac {1}{2} = 1$

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Single Correct Hard Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 109

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