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paglia

  • 2 years ago

sin^3x + cos^4x

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  1. LagrangeSon678
    • 2 years ago
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    what is the problem here?

  2. paglia
    • 2 years ago
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    how to integrate, sorry!

  3. LagrangeSon678
    • 2 years ago
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    Well, if the power of sine is odd and positive , what you want to do is lop of one of the sine factor, put it to the right of the rest of the expression and convert the remaining (even) sine factors to cosines withe the Pythagorean identity , and then integrate with the substitution method where u=cos(x)

  4. LagrangeSon678
    • 2 years ago
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    |dw:1325279411322:dw|

  5. paglia
    • 2 years ago
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    oh, ok, then, thanks.

  6. LagrangeSon678
    • 2 years ago
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    hold on let me finish it for you: \[\int sin^3(x)cos^4(x)dx =\int sin^2(x)cos^4(x)sin(x)dx\]

  7. paglia
    • 2 years ago
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    :D

  8. LagrangeSon678
    • 2 years ago
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    \[\int sin^2(x)cos^4(x)sin(x)dx=\int (1-cos^2(x))cos^4(x)sin(x)dx\]

  9. LagrangeSon678
    • 2 years ago
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    \[\int (1-cos^2(x))cos^4(x)sin(x)dx=\int (cos^4(x)-cos^6(x))sin(x)dx\]

  10. LagrangeSon678
    • 2 years ago
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    \[u=cos(x)\rightarrow \frac{du}{dx}=-sin(x)\rightarrow du=-sin(x)dx\rightarrow \frac{-du}{sin(x)}=dx\]

  11. LagrangeSon678
    • 2 years ago
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    \[\int (cos^4(x)-cos^6(x))sin(x)dx=-\int (u^4-u^6)du\]

  12. LagrangeSon678
    • 2 years ago
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    \[=-\int (u^4-u^6)du=-\frac{1}{5}u^5+\frac{1}{7}u^7+C\]

  13. LagrangeSon678
    • 2 years ago
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    \[=\frac{1}{7}cos^7(x)-\frac{1}{5}cos^5(x)+C\]

  14. LagrangeSon678
    • 2 years ago
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    And thats your answer

  15. paglia
    • 2 years ago
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    so simple. thanks!

  16. LagrangeSon678
    • 2 years ago
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    well, its simple once you see it

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