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Question Number 147837 by peter frank last updated on 23/Jul/21

Evaluate ∫((sin^8 θ−cos^8 θ)/(1−2sin^2 θcos^2 θ)) dθ

$${Evaluate} \\ $$$$\int\frac{\mathrm{sin}\:^{\mathrm{8}} \theta−\mathrm{cos}\:^{\mathrm{8}} \theta}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} \theta\mathrm{cos}\:^{\mathrm{2}} \theta}\:{d}\theta \\ $$$$ \\ $$

Answered by liberty last updated on 24/Jul/21

sin^8 x−cos^8 x=(sin^4 x+cos^4 x)(sin^4 x−cos^4 x) =(1−2sin^2 xcos^2 x)(sin^2 x−cos^2 x) I=∫(((1−2sin^2 x cos^2 x)(sin^2 x−cos^2 x))/(1−2sin^2 x cos^2 x)) dx = ∫ −cos 2x dx =−(1/2)sin 2x + c =−sin x cos x + c

$$\mathrm{sin}\:^{\mathrm{8}} {x}−\mathrm{cos}\:^{\mathrm{8}} {x}=\left(\mathrm{sin}\:^{\mathrm{4}} {x}+\mathrm{cos}\:^{\mathrm{4}} {x}\right)\left(\mathrm{sin}\:^{\mathrm{4}} {x}−\mathrm{cos}\:^{\mathrm{4}} {x}\right) \\ $$$$=\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\mathrm{cos}\:^{\mathrm{2}} {x}\right)\left(\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{cos}\:^{\mathrm{2}} {x}\right) \\ $$$${I}=\int\frac{\left(\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}\right)\left(\mathrm{sin}\:^{\mathrm{2}} {x}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)}{\mathrm{1}−\mathrm{2sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}}\:{dx} \\ $$$$=\:\int\:−\mathrm{cos}\:\mathrm{2}{x}\:{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\mathrm{2}{x}\:+\:{c} \\ $$$$=−\mathrm{sin}\:{x}\:\mathrm{cos}\:{x}\:+\:{c}\: \\ $$

Commented by peter frank last updated on 24/Jul/21

thank you

$${thank}\:{you} \\ $$

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