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Question Number 167438 by cortano1 last updated on 16/Mar/22

  ∫ ((sin 4x)/(4sin^4 x−4sin^2 x+1)) dx=?

$$\:\:\int\:\frac{\mathrm{sin}\:\mathrm{4x}}{\mathrm{4sin}\:^{\mathrm{4}} \mathrm{x}−\mathrm{4sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{1}}\:\mathrm{dx}=? \\ $$

Answered by MJS_new last updated on 16/Mar/22

((sin 4x)/(1−4sin^2  x +4sin^4  x))=((2sin 4x)/(1+cos 4x))=((2sin 2x cos 2x)/(cos^2  2x))=  =2tan 2x  2∫tan 2x dx=−ln ∣cos 2x∣ +C

$$\frac{\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{1}−\mathrm{4sin}^{\mathrm{2}} \:{x}\:+\mathrm{4sin}^{\mathrm{4}} \:{x}}=\frac{\mathrm{2sin}\:\mathrm{4}{x}}{\mathrm{1}+\mathrm{cos}\:\mathrm{4}{x}}=\frac{\mathrm{2sin}\:\mathrm{2}{x}\:\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{cos}^{\mathrm{2}} \:\mathrm{2}{x}}= \\ $$$$=\mathrm{2tan}\:\mathrm{2}{x} \\ $$$$\mathrm{2}\int\mathrm{tan}\:\mathrm{2}{x}\:{dx}=−\mathrm{ln}\:\mid\mathrm{cos}\:\mathrm{2}{x}\mid\:+{C} \\ $$

Answered by cortano1 last updated on 17/Mar/22

 Another way   Z=∫ ((sin 4x)/((2sin^2 x −1)^2 )) dx    Z= ∫ ((sin 4x )/((−cos 2x)^2 )) dx   Z=∫ ((sin 4x)/((((1+cos 4x)/2)))) dx   Z=−(1/2)∫ ((d(1+cos 4x))/(1+cos 4x))   Z=−(1/2)ln ∣1+cos 4x∣ + c

$$\:\mathrm{Another}\:\mathrm{way} \\ $$$$\:\mathrm{Z}=\int\:\frac{\mathrm{sin}\:\mathrm{4x}}{\left(\mathrm{2sin}^{\mathrm{2}} \mathrm{x}\:−\mathrm{1}\right)^{\mathrm{2}} }\:\mathrm{dx}\: \\ $$$$\:\mathrm{Z}=\:\int\:\frac{\mathrm{sin}\:\mathrm{4x}\:}{\left(−\mathrm{cos}\:\mathrm{2x}\right)^{\mathrm{2}} }\:\mathrm{dx} \\ $$$$\:\mathrm{Z}=\int\:\frac{\mathrm{sin}\:\mathrm{4x}}{\left(\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{4x}}{\mathrm{2}}\right)}\:\mathrm{dx} \\ $$$$\:\mathrm{Z}=−\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{d}\left(\mathrm{1}+\mathrm{cos}\:\mathrm{4x}\right)}{\mathrm{1}+\mathrm{cos}\:\mathrm{4x}} \\ $$$$\:\mathrm{Z}=−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{1}+\mathrm{cos}\:\mathrm{4x}\mid\:+\:\mathrm{c}\: \\ $$$$ \\ $$

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