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Miscellaneous Exercise 4 · Q76

Q.Evaluate: ∫0π/4cos⁡2x1+cos⁡2x+sin⁡2x dx\int_0^{\pi/4} \dfrac{\cos 2x}{1+\cos 2x+\sin 2x}\,dx

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1+cos⁡2x=2cos⁡2x1+\cos2x=2\cos^2x, sin⁡2x=2sin⁡xcos⁡x\sin2x=2\sin x\cos x, so the denominator is 2cos⁡x(cos⁡x+sin⁡x)2\cos x(\cos x+\sin x); and cos⁡2x=(cos⁡x−sin⁡x)(cos⁡x+sin⁡x)\cos2x=(\cos x-\sin x)(\cos x+\sin x).

cos⁡2x1+cos⁡2x+sin⁡2x=cos⁡x−sin⁡x2cos⁡x=12−12tan⁡x.\frac{\cos2x}{1+\cos2x+\sin2x}=\frac{\cos x-\sin x}{2\cos x}=\frac12-\frac12\tan x. …

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