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anonymous

  • 5 years ago

int_{0}^{3} x/(sqrt{x^2+16}

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  1. anonymous
    • 5 years ago
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    let\[x =\tan \theta\]

  2. anonymous
    • 5 years ago
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    :) thx

  3. anonymous
    • 5 years ago
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    On closer inspection, you don't need trig sub, u sub would suffice. I never noticed until after I worked it out myself. But hey, good practice for trig sub. When you get it on the test you going to knock it out of the park.

  4. amistre64
    • 5 years ago
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    shouldnt x = 4tan(t)? that way; when you square it you get: sqrt(16 tan^2(t) + 16) = sqrt(16(tan^2+1))) = 4sec^2

  5. amistre64
    • 5 years ago
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    err.... 4 sec :)

  6. anonymous
    • 5 years ago
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    Yes. I think he picked that up.

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