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Eherre1989

  • 4 years ago

can someone solve this for me without changing the limits of integration?

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  1. Eherre1989
    • 4 years ago
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    \[\int\limits_{1}^{3}x \sqrt{1-(x-2)^2}dx\]

  2. Eherre1989
    • 4 years ago
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    step by step please

  3. Eherre1989
    • 4 years ago
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    it has to be a trig substitution i just get stuck when i have to integrate the sins

  4. Mr.Math
    • 4 years ago
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    Write what you have so far.

  5. Eherre1989
    • 4 years ago
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    \[\int\limits_{}^{}(\sin \theta-\sin^3\theta-2\sin^2\theta+2)d \theta\]

  6. Eherre1989
    • 4 years ago
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    i need that integrated i cant seem to figure out the middle two

  7. Eherre1989
    • 4 years ago
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    i feel as if i should do by parts

  8. Mr.Math
    • 4 years ago
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    Break it down into four integrals. I'm sure you know how to integrate \(\sin\theta\) and \(2\).

  9. Mr.Math
    • 4 years ago
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    For \(\sin^3\theta\), write \(\sin^3\theta=\sin\theta-\cos^2\theta\sin\theta\), and then the integral of \(\sin\theta\) is easy. Substitute \(u=\cos\theta\) for the integration of \(-\cos^2\theta\sin\theta\).

  10. Eherre1989
    • 4 years ago
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    i dont remember any trig identities

  11. Mr.Math
    • 4 years ago
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    \(\sin^2\theta=\frac{1-\cos(2\theta)}{2}\).

  12. Mr.Math
    • 4 years ago
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    Sorry it should be \(-2\sin^2\theta=\cos(2\theta)-1\).

  13. Eherre1989
    • 4 years ago
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    oh ok thanks

  14. Mr.Math
    • 4 years ago
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    You're welcome.

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