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Question Number 79763 by mathmax by abdo last updated on 28/Jan/20

calculate ∫_0 ^π {cos^8 x +sin^8 x}dx

$${calculate}\:\int_{\mathrm{0}} ^{\pi} \left\{{cos}^{\mathrm{8}} {x}\:+{sin}^{\mathrm{8}} {x}\right\}{dx} \\ $$

Commented by john santu last updated on 28/Jan/20

cos^8 x+sin^8 x=(sin^4 x)^2 +(cos^4 x)^2 = (sin^4 x+cos^4 x)^2 −2sin^4 xcos^4 x = {(sin^2 x+cos^2 x)^2 −2sin^2 xcos^2 x}^2 −(1/2)sin^2 (2x) = {1−(1/2)sin^2 (2x)}^2 −(1/2)sin^2 (2x) =1−(3/2)sin^2 (2x)+(1/4)sin^4 (2x) =1−(3/2)((1/2)−(1/2)cos (4x))+(1/4){(1/2)−(1/2)cos (4x)}^2 =(1/4)+(3/4)cos (4x)+(1/4){(1/4)−(1/2)cos (4x)+(1/4)cos^2 (4x)} = (5/(16))+(5/8)cos (4x)+(1/(16))((1/2)+(1/2)cos (8x)) = ((11)/(32))+(5/8)cos (4x)+(1/(32))cos (8x)

$$\mathrm{cos}\:^{\mathrm{8}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{8}} \mathrm{x}=\left(\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}\right)^{\mathrm{2}} +\left(\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\right)^{\mathrm{2}} \\ $$$$=\:\left(\mathrm{sin}\:^{\mathrm{4}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}\right)^{\mathrm{2}} −\mathrm{2sin}\:^{\mathrm{4}} \mathrm{xcos}\:^{\mathrm{4}} \mathrm{x} \\ $$$$=\:\left\{\left(\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}\right)^{\mathrm{2}} −\mathrm{2sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}\right\}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right) \\ $$$$=\:\left\{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\right\}^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{4}} \left(\mathrm{2x}\right) \\ $$$$=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{4x}\right)\right)+\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{4x}\right)\right\}^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{cos}\:\left(\mathrm{4x}\right)+\frac{\mathrm{1}}{\mathrm{4}}\left\{\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{4x}\right)+\frac{\mathrm{1}}{\mathrm{4}}\mathrm{cos}\:^{\mathrm{2}} \left(\mathrm{4x}\right)\right\} \\ $$$$=\:\frac{\mathrm{5}}{\mathrm{16}}+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{cos}\:\left(\mathrm{4x}\right)+\frac{\mathrm{1}}{\mathrm{16}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\left(\mathrm{8x}\right)\right) \\ $$$$=\:\frac{\mathrm{11}}{\mathrm{32}}+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{cos}\:\left(\mathrm{4x}\right)+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{cos}\:\left(\mathrm{8x}\right) \\ $$

Commented by john santu last updated on 28/Jan/20

∫_0 ^π (((11)/(32))+(5/8)cos (4x)+(1/(32))cos (8x))dx= ((11x)/(32))+(5/(32))sin (4x)+(1/(256))sin (8x)∣_0 ^π = ((11π)/(32)) ★

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\:\left(\frac{\mathrm{11}}{\mathrm{32}}+\frac{\mathrm{5}}{\mathrm{8}}\mathrm{cos}\:\left(\mathrm{4x}\right)+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{cos}\:\left(\mathrm{8x}\right)\right)\mathrm{dx}= \\ $$$$\frac{\mathrm{11x}}{\mathrm{32}}+\frac{\mathrm{5}}{\mathrm{32}}\mathrm{sin}\:\left(\mathrm{4x}\right)+\frac{\mathrm{1}}{\mathrm{256}}\mathrm{sin}\:\left(\mathrm{8x}\right)\underset{\mathrm{0}} {\overset{\pi} {\mid}} \\ $$$$=\:\frac{\mathrm{11}\pi}{\mathrm{32}}\:\bigstar \\ $$

Answered by Henri Boucatchou last updated on 28/Jan/20

we have cos^8 x+sin^8 x=1−(8/8)sin^2 2x =cos^2 2x=(1/4)(e^(i2x) +e^(−i2x) )^2 =(1/4)[(e^(i4x) +e^(−i4x) )+2]=(1/4)(2cos4x+2) ⇒ ∫_0 ^( π) (cos^8 x+sin^8 x)dx=(1/2)[(1/4)sin4x+2x]_0 ^π =(1/2)(2π) =π

$${we}\:\:{have}\:\:{cos}^{\mathrm{8}} {x}+{sin}^{\mathrm{8}} {x}=\mathrm{1}−\frac{\mathrm{8}}{\mathrm{8}}{sin}^{\mathrm{2}} \mathrm{2}{x} \\ $$$$={cos}^{\mathrm{2}} \mathrm{2}{x}=\frac{\mathrm{1}}{\mathrm{4}}\left({e}^{{i}\mathrm{2}{x}} +{e}^{−{i}\mathrm{2}{x}} \right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left[\left({e}^{{i}\mathrm{4}{x}} +{e}^{−{i}\mathrm{4}{x}} \right)+\mathrm{2}\right]=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{cos}\mathrm{4}{x}+\mathrm{2}\right) \\ $$$$\Rightarrow\:\int_{\mathrm{0}} ^{\:\pi} \left({cos}^{\mathrm{8}} {x}+{sin}^{\mathrm{8}} {x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left[\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{4}{x}+\mathrm{2}{x}\right]_{\mathrm{0}} ^{\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}\pi\right) \\ $$$$=\pi \\ $$

Commented by john santu last updated on 28/Jan/20

how to get sin^8 x+cos^8 x=1−(8/8)sin^2 2x? it impossible

$$\mathrm{how}\:\mathrm{to}\:\mathrm{get}\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{x}+\mathrm{cos}\:^{\mathrm{8}} \mathrm{x}=\mathrm{1}−\frac{\mathrm{8}}{\mathrm{8}}\mathrm{sin}\:^{\mathrm{2}} \mathrm{2x}? \\ $$$$\mathrm{it}\:\mathrm{impossible} \\ $$

Commented by john santu last updated on 28/Jan/20

a^4 +b^4 =(a+b)^4 −2ab(2a^2 −3ab−2b^2 ) put a=cos^2 x, b = sin^2 x cos^8 x+sin^8 x=1−(1/2).4sin^2 xcos^2 x(2(cos^2 x−sin^2 x)−(3/4).4sin^2 xcos^2 x) = 1−(1/2)sin^2 (2x)(2cos (2x)−(3/4)sin^2 (2x)) = 1−(1/4)sin (4x)sin (2x)+(3/8)sin^4 (2x)

$$\mathrm{a}^{\mathrm{4}} +\mathrm{b}^{\mathrm{4}} =\left(\mathrm{a}+\mathrm{b}\right)^{\mathrm{4}} −\mathrm{2ab}\left(\mathrm{2a}^{\mathrm{2}} −\mathrm{3ab}−\mathrm{2b}^{\mathrm{2}} \right) \\ $$$$\mathrm{put}\:\mathrm{a}=\mathrm{cos}\:^{\mathrm{2}} \mathrm{x},\:\mathrm{b}\:=\:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x} \\ $$$$\mathrm{cos}\:^{\mathrm{8}} \mathrm{x}+\mathrm{sin}\:^{\mathrm{8}} \mathrm{x}=\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}.\mathrm{4sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}\left(\mathrm{2}\left(\mathrm{cos}\:^{\mathrm{2}} \mathrm{x}−\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}\right)−\frac{\mathrm{3}}{\mathrm{4}}.\mathrm{4sin}\:^{\mathrm{2}} \mathrm{xcos}\:^{\mathrm{2}} \mathrm{x}\right) \\ $$$$=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\left(\mathrm{2cos}\:\left(\mathrm{2x}\right)−\frac{\mathrm{3}}{\mathrm{4}}\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2x}\right)\right) \\ $$$$=\:\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\left(\mathrm{4x}\right)\mathrm{sin}\:\left(\mathrm{2x}\right)+\frac{\mathrm{3}}{\mathrm{8}}\mathrm{sin}\:^{\mathrm{4}} \left(\mathrm{2x}\right) \\ $$

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