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Math Help - integral (trig)

  1. #1
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    integral (trig)

    Find the integral of:

    csc(x) * dx
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  2. #2
    Senior Member Peritus's Avatar
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    csc(x) = cos(x) / (cos(x)^2)
    u = sin(x)
    du = cos(x)dx

    now using the trigonometric identity sin^2(x) + cos^2(x) = 1
    we arrive at the following equivalent integral

    du/(1-u^2)

    which can be written as:

    0.5*[1/(1-u) + 1/(1+u)]du

    integrating we get:

    0.5*[ln(1+u) - ln(1-u)] = 0.5*ln[(1+u)/(1-u)]
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  3. #3
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Peritus View Post
    csc(x) = cos(x) / (cos(x)^2)
    I'd re-think that.

    --

    About the integral, multiply it by \csc x-\cot x.
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  4. #4
    GAMMA Mathematics
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    Quote Originally Posted by DINOCALC09 View Post
    Find the integral of:

    csc(x) * dx
    My first method is a bit long, but I propose double substitution of:

    u=sin(x), so csc(x) = \frac{1}{u}

    You will then have du=cos(x)dx
    To solve for cosine in terms of u, use the sine, cosine relationship:

    cos^2(x)+sin^2(x)=1
    cos^2(x)=1-u^2
    cos(x)=\sqrt{1-u^2}

    You will then have an integrand of \frac{1}{u\sqrt{1-u^2}}

    To simplify that, you need to make the substitution: v=1-u^2
    dv=-2udu
    u=\sqrt{v-1}

    The resulting integrand becomes:

    \frac{-2}{\sqrt{v}} = -2v^{-1/2}

    Integrating that, you get: -4v^{1/2}

    Substitution of v=1-u^2 leads to...

    -4{(1-u^2)}^{1/2}

    Substitution of u=sin(x) leads to...

    -4{(1-sin^2(x))}^{1/2} = -4\sqrt{cos(x)}

    Hmmm, doesn't look like the answer I should get?
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  5. #5
    Senior Member Peritus's Avatar
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    sorry my mistake, but the solution changes only slightly

    csc(x) = sin(x) / (sin(x)^2)
    u = cos(x)
    du = -sin(x)dx

    now using the trigonometric identity sin^2(x) + cos^2(x) = 1
    we arrive at the following equivalent integral

    -du/(1-u^2)

    which can be written as:

    -0.5*[1/(1-u) + 1/(1+u)]du

    integrating we get:

    -0.5*[ln(1+u) - ln(1-u)] = 0.5*ln[(1-u)/(1+u)] + const =
    0.5*ln[(1-cos(x))/(1+cos(x))] + const
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  6. #6
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    The book has the answer as:

    -ln|csc(x)+cot(x)| + c

    The only problem is that it doesn't show me how to get there.
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  7. #7
    Senior Member Peritus's Avatar
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    -ln[csc(x)+cot(x)]=-ln[(1/sin(x) + cos(x)/sin(x)] = ln[sin(x)/(1+cos(x))]

    <=> 0.5*ln[sin^2(x) / ((1+cos(x))^2)] =
    = 0.5*ln[(1-cos^2(x)) / ((1+cos(x))^2)] =
    = 0.5*ln{[(1-cos(x))*(1+cos(x))] / ((1+cos(x))^2)} = 0.5*ln[(1-cos(x))/(1+cos(x))]
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