Mathematics

# Prove that $\int\limits_0^{\tfrac {32\pi} 3} {\sqrt {1 + \cos 2x} \,dx = \sqrt {\frac{3}{2}} }$

##### SOLUTION
$\displaystyle\int_{0}^{\frac{32\pi}{3}}{\sqrt{1+\cos{2x}}dx}$
$=\displaystyle\int_{0}^{\frac{32\pi}{3}}{\sqrt{2{\cos}^{2}{x}}dx}$
$=\sqrt{2}\displaystyle\int_{0}^{\frac{32\pi}{3}}{\cos{x}dx}$
$=\sqrt{2}\left[\sin{x}\right]_{0}^{\frac{32\pi}{3}}$
$=\sqrt{2}\left[\sin{\dfrac{32\pi}{3}}-\sin{0}\right]$
$=\sqrt{2}\left[\sin{\left(10\pi+\dfrac{2\pi}{3}\right)}-0\right]$
$=\sqrt{2}\sin{\dfrac{2\pi}{3}}$
$=\sqrt{2}\sin{\left(\pi-\dfrac{\pi}{3}\right)}$
$=\sqrt{2}\sin{\dfrac{\pi}{3}}$
$=\sqrt{2}\times\dfrac{\sqrt{3}}{2}$
$=\sqrt{2}\times\dfrac{\sqrt{3}}{2}\times\dfrac{\sqrt{2}}{\sqrt{2}}$
$=\sqrt{\dfrac{3}{2}}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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