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Denebel

  • 4 years ago

How do I integrate this?

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  1. Denebel
    • 4 years ago
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    |dw:1326787674889:dw|

  2. Denebel
    • 4 years ago
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    I tried using u=x? What should u be? |dw:1326787713147:dw|

  3. imperialist
    • 4 years ago
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    Hint: what do you get when you take the derivative of \[\cot x = \frac{\cos x}{\sin x}\]

  4. Denebel
    • 4 years ago
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    Derivative of cot x? =-(arccsc x)^2

  5. imperialist
    • 4 years ago
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    Not arccsc, but rather \[-\csc^2 x = \frac{-1}{\sin^2x}\]

  6. DHASHNI
    • 4 years ago
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    |dw:1326788429512:dw|

  7. DHASHNI
    • 4 years ago
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    |dw:1326788552348:dw|

  8. Denebel
    • 4 years ago
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    Oh yes, my bad. So.. |dw:1326788533509:dw| ?

  9. Denebel
    • 4 years ago
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    Where did the neg. come from?

  10. imperialist
    • 4 years ago
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    \[\int \frac{3 dx}{\sin^2x}=3\int \csc^2xdx = 3(-\cot x)+C=-3\cot x + C\]

  11. DHASHNI
    • 4 years ago
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    if we integrate cosec^2 x =>-cot x

  12. Denebel
    • 4 years ago
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    Thanks!

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