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alfers101

  • 2 years ago

integratation of power of sines and cosines: ∫ sin^4 xcos^2xdx

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  1. alfers101
    • 2 years ago
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    this is calculus

  2. marsss
    • 2 years ago
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    \[-\frac{ \cos ^{5}x }{ 5 }+\frac{ \cos ^{7}x }{ 7 }+c\]

  3. alfers101
    • 2 years ago
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    how come? what case did you used? is it case 2?

  4. marsss
    • 2 years ago
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    bak reyiz, cos^2 x=1-sin^2 x. biliyon mu bunu??

  5. alfers101
    • 2 years ago
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    what? im sorry i cant understant your language, could you speak in english please?

  6. alfers101
    • 2 years ago
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    can someone please help me? :(

  7. Callisto
    • 2 years ago
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    I think it's not like what marsss did there... \[∫ sin^4 xcos^2xdx\]\[=∫ sin^4 x(1-sin^2x)dx\]\[=∫ sin^4x-sin^6xdx\]\[=∫ (sin^2x)^2dx-∫(sin^2x)^3dx\]\[=\frac{1}{4}∫ (1-cos2x)^2dx-\frac{1}{8}∫(1-cos2x)^3dx\]\[=\frac{1}{4}∫ (1-2cos2x+cos^22x)dx-\frac{1}{8}∫(1-3cos2x+3cos^22x-cos^32x)dx\] Reduce the power of cosine into 1 using doucle angle formula, except for cos^3(2x). Good luck :|

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