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Question Number 141249 by bramlexs22 last updated on 17/May/21

Commented by Rozix last updated on 17/May/21

$$kindlydropmeyourwhatsappnumbersifinterestedinhavingamathsgroup$$

Answered by MJS_new last updated on 17/May/21

$$-\underset{0}{\stackrel{\pi /2}{\int }}\frac{1-\cos 2x}{3+\cos 4x}dx=$$$$[t=\tan x\to dx={\cos }^{2}xdt]$$$$=-\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{t^2}{t^4+1}dt=$$$$={[\frac{\sqrt{2}}{16}\ln \frac{t^2+\sqrt{2}t+1}{t^2-\sqrt{2}t+1}-\frac{\sqrt{2}}{8}(\arctan (\sqrt{2}t-1)+\arctan (\sqrt{2}t+1))]}_0^{\mathrm{\infty }}=$$$$=-\frac{\sqrt{2}\pi }{8}$$

Commented by MJS_new last updated on 17/May/21

$$-\frac{\sqrt{2}\pi }{8}=-\frac{\sqrt{2}\pi \sqrt{2}}{8\sqrt{2}}=-\frac{2\pi }{8\sqrt{2}}=-\frac{\pi }{4\sqrt{2}}$$

Answered by bemath last updated on 17/May/21

$$I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{\cos 2x-1}{2\cos {}^{2}2x+2}dx$$$$2I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{\cos 2x-1}{\cos {}^{2}2x+1}$$$$letu=\frac{\pi }{2}-x$$$$2I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{-\cos 2x-1}{\cos {}^{2}2x+1}dx$$$$4I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{-2}{\cos {}^{2}2x+1}dx$$$$-2I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sec {}^{2}2x}{\sec {}^{2}2x+1}dx$$$$-2I=\underset{0}{\stackrel{\pi /2}{\int }}\frac{\sec {}^{2}2x}{\tan {}^{2}2x+{\left(\sqrt{2}\right)}^2}dx$$$$-2I=\frac{1}{2}\underset{0}{\stackrel{\mathrm{\infty }}{\int }}\frac{dx}{x^2+{\left(\sqrt{2}\right)}^2}$$$$-2I=\frac{1}{2}.\frac{1}{\sqrt{2}}{\left[\arctan \left(\frac{x}{\sqrt{2}}\right)\right]}_0^{\mathrm{\infty }}$$$$I=-\frac{1}{4\sqrt{2}}.\frac{\pi }{2}=-\frac{\pi \sqrt{2}}{8}$$

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