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0-1-sin-x-cos-2-x-cos-2-pi-5-cos-2-x-cos-2-2pi-5-sin-5x-dx-




Question Number 177906 by cortano1 last updated on 11/Oct/22
  ∫_0 ^1  ((sin x(cos^2 x−cos^2 (π/5))(cos^2 x−cos^2 ((2π)/5)))/(sin 5x)) dx =?
$$\:\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{\mathrm{sin}\:\mathrm{x}\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{5}}\right)\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{5}}\right)}{\mathrm{sin}\:\mathrm{5x}}\:\mathrm{dx}\:=? \\ $$
Answered by Frix last updated on 11/Oct/22
((sin x(cos^2 x−cos^2 (π/5))(cos^2 x−cos^2 ((2π)/5)))/(sin 5x))=(1/(16))  ∫_0 ^1 (dx/(16))=(1/(16))
$$\frac{\mathrm{sin}\:\mathrm{x}\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{5}}\right)\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{cos}\:^{\mathrm{2}} \frac{\mathrm{2}\pi}{\mathrm{5}}\right)}{\mathrm{sin}\:\mathrm{5x}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\frac{{dx}}{\mathrm{16}}=\frac{\mathrm{1}}{\mathrm{16}} \\ $$
Commented by Frix last updated on 11/Oct/22
sin x =s  ⇒  cos^2  x −cos^2  (π/5) =−s^2 +(5/8)−((√5)/8)  cos^2  x −cos^2  ((2π)/5) =−s^2 +(5/8)+((√5)/8)  sin 5x =16s^5 −20s^3 +5s  the rest is easy
$$\mathrm{sin}\:{x}\:={s} \\ $$$$\Rightarrow \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{cos}^{\mathrm{2}} \:\frac{\pi}{\mathrm{5}}\:=−{s}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{8}}−\frac{\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}\:−\mathrm{cos}^{\mathrm{2}} \:\frac{\mathrm{2}\pi}{\mathrm{5}}\:=−{s}^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{8}}+\frac{\sqrt{\mathrm{5}}}{\mathrm{8}} \\ $$$$\mathrm{sin}\:\mathrm{5}{x}\:=\mathrm{16}{s}^{\mathrm{5}} −\mathrm{20}{s}^{\mathrm{3}} +\mathrm{5}{s} \\ $$$$\mathrm{the}\:\mathrm{rest}\:\mathrm{is}\:\mathrm{easy} \\ $$

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