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Question Number 108507 by Eric002 last updated on 17/Aug/20

$${\int }_0^{\frac{\pi }{6}}\sqrt{\frac{3cos2x-1}{{cos}^2\left(x\right)}}dx$$

Answered by Sarah85 last updated on 18/Aug/20

$$\underset{0}{\stackrel{\frac{\pi }{6}}{\int }}\sqrt{\frac{3\cos 2x-1}{{\cos }^{2}x}}dx=\sqrt{2}\underset{0}{\stackrel{\frac{\pi }{6}}{\int }}\sqrt{\frac{3\cos 2x-1}{1+\cos 2x}}dx$$$$\mathrm{first}\mathrm{step}$$$$t=\tan x\iff x=\arctan t\iff dx=\frac{dt}{t^2+1}$$$$\sqrt{2}\underset{0}{\stackrel{\frac{\sqrt{3}}{3}}{\int }}\frac{\sqrt{1-2t^2}}{t^2+1}dt$$$$\mathrm{second}\mathrm{step}$$$$t=\frac{\sqrt{2}}{2}\sin u\iff u=\arcsin \sqrt{2}t\iff dt=\frac{\sqrt{2}}{2}\sqrt{1-2t^2}$$$$2\underset{0}{\stackrel{\arcsin \frac{\sqrt{6}}{3}}{\int }}\frac{{\cos }^{2}u}{2+{\sin }^{2}u}du$$$$\mathrm{third}\mathrm{step}$$$$v=\tan u\iff u=\arctan v\iff du=\frac{dv}{v^2+1}$$$$2\underset{0}{\stackrel{\sqrt{2}}{\int }}\frac{1}{(v^2+1)(3v^2+2)}dv=$$$$=6\underset{0}{\stackrel{\sqrt{2}}{\int }}\frac{1}{3v^2+2}dv-2\underset{0}{\stackrel{\sqrt{2}}{\int }}\frac{1}{v^2+1}dv=$$$$={[\sqrt{6}\arctan \frac{\sqrt{6}v}{2}-2\arctan v]}_0^{\sqrt{2}}=$$$$=\frac{\sqrt{6}}{3}\pi -2\arctan \sqrt{2}$$

Commented by bobhans last updated on 18/Aug/20

$$1+\cos 2x=1+2\cos {}^{2}x-1=2\cos {}^{2}x$$$$buttheoriginalequationindenumerator$$$$is\cos {}^{2}x$$

Commented by Sarah85 last updated on 18/Aug/20

$$\mathrm{typo},\mathrm{I}\mathrm{corrected}\mathrm{it}$$

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