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Thread: Integral

  1. #1
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    Integral

    How do I solve using substitution?

    $\displaystyle \int\sin4xtanxdx$

    Thanks for your help.
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  2. #2
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    $\displaystyle \int\sin4x\tan{x}~dx=\int\frac{\sin4x\sin{x}}{\cos {x}}~dx=\frac12\int\frac{\cos3x-\cos5x}{\cos{x}}~dx$
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    $\displaystyle \int\sin4x\tan{x}~dx=\int\frac{\sin4x\sin{x}}{\cos {x}}~dx=\frac12\int\frac{\cos3x-\cos5x}{\cos{x}}~dx$
    Thanks for the help. Can you explain the last step on how you got that...and...

    I copied the problem incorrect it's:
    $\displaystyle \int\sec4xtanxdx$
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by xxpl510xx View Post
    Thanks for the help. Can you explain the last step on how you got that
    Consider the following two identities:
    $\displaystyle cos(a - b) = cos(a)cos(b) + sin(a)sin(b)$
    $\displaystyle cos(a + b) = cos(a)cos(b) - sin(a)sin(b)$

    Subtract the bottom from the top:
    $\displaystyle cos(a - b) - cos(a + b) = 2sin(a)sin(b)$

    Thus
    $\displaystyle sin(a)sin(b) = \frac{1}{2}(cos(a - b) - cos(a + b))$

    Thus
    $\displaystyle sin(4x)sin(x) = \frac{1}{2}(cos(4x - x) - cos(4x + x))$

    $\displaystyle sin(4x)sin(x) = \frac{1}{2}(cos(3x) - cos(5x))$

    -Dan
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  5. #5
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    Quote Originally Posted by xxpl510xx View Post
    Thanks for the help. Can you explain the last step on how you got that...and...

    I copied the problem incorrect it's:
    $\displaystyle \int\sec4xtanxdx$
    I love jokes. This one rocks!!
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  6. #6
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    Quote Originally Posted by xxpl510xx View Post
    Thanks for the help. Can you explain the last step on how you got that...and...

    I copied the problem incorrect it's:
    $\displaystyle \int\sec4xtanxdx$
    I bet also that the new one is supposed to be
    INT.[(secX)^4 tanX] dX

    So,
    = INT.[(secX)^3](secX tanX dX)
    = (1/4)(secX)^4 +C -------------------answer.
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