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Exercise 11.13 · Q12

Q.∫sin⁡8x−cos⁡8x1−2sin⁡2xcos⁡2x dx\displaystyle\int \dfrac{\sin^8 x-\cos^8 x}{1-2\sin^2x\cos^2x}\,dx is

(1) 12sin⁡2x+c\dfrac12\sin 2x+c
(2) −12sin⁡2x+c-\dfrac12\sin 2x+c
(3) 12cos⁡2x+c\dfrac12\cos 2x+c
(4) −12cos⁡2x+c-\dfrac12\cos 2x+c
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Factor the numerator and recognise it shares the denominator's factor sin⁡4x+cos⁡4x\sin^4x+\cos^4x.

Step 1. Numerator =(sin⁡4x−cos⁡4x)(sin⁡4x+cos⁡4x)=(sin⁡2x−cos⁡2x)(sin⁡4x+cos⁡4x)=−cos⁡2x (sin⁡4x+cos⁡4x)=(\sin^4x-\cos^4x)(\sin^4x+\cos^4x)=(\sin^2x-\cos^2x)(\sin^4x+\cos^4x)=-\cos2x\,(\sin^4x+\cos^4x).

Step 2. sin⁡4x+cos⁡4x=(sin⁡2x+cos⁡2x)2−2sin⁡2xcos⁡2x=1−2sin⁡2xcos⁡2x\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x, exactly the denominator. …

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