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Exercise 4.2 · Q23

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

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sin⁡4x+cos⁡4x=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x=1−12sin⁡22x\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac12\sin^22x. Let t=cos⁡2x, dt=−2sin⁡2x dxt=\cos2x,\ dt=-2\sin2x\,dx, so sin⁡22x=1−t2\sin^22x=1-t^2 and the denominator becomes 1+t22\frac{1+t^2}2.

∫sin⁡2x dx1−12(1−t2)=∫−dt/2(1+t2)/2=−∫dt1+t2.\int\frac{\sin2x\,dx}{1-\frac12(1-t^2)}=\int\frac{-dt/2}{(1+t^2)/2}=-\int\frac{dt}{1+t^2}. …

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