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
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