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Question Number 53894 by muhsangsll last updated on 27/Jan/19

$$\underset{\pi /6}{\stackrel{5\pi /6}{\int }}\sqrt{4-4{\sin }^{2}t}dt=$$

Answered by byaw last updated on 27/Jan/19

$${\int }_{\pi /6}^{5\pi /6}\sqrt{4(1-{sin}^{2}t)}dt$$$${\int }_{\pi /6}^{5\pi /6}\sqrt{4\left({cos}^{2}t\right)}dt$$$${\int }_{\pi /6}^{5\pi /6}2costdt$$$$2{\int }_{\pi /6}^{5\pi /6}costdt$$$$\text{Missing left or extra right}$$$$2[sin\left(5\pi /6\right)-sin\left(\pi /6\right)]$$$$2(1/2-1/2)$$$$0$$

Answered by ajfour last updated on 27/Jan/19

$$=2{\int }_{\pi /6}^{\pi -\pi /6}\mid \cos t\mid dt$$$$=2{\int }_{\pi /6}^{\pi /2}\cos tdt-2{\int }_{\pi /2}^{\pi -\pi /6}\cos tdt$$$$=2\left(\sin t\right){\mid }_{\pi /6}^{\pi /2}-2\left(\sin t\right){\mid }_{\pi /6}^{\pi -\pi /6}$$$$=2(1-\frac{1}{2})-2\times 0=1.$$

Commented by peter frank last updated on 27/Jan/19

$$\mathrm{p}l\mathrm{ease}\mathrm{help}53907$$

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