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Math Help - Integrate 1/[5+4cos(x)] with limits?

  1. #1
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    Integrate 1/[5+4cos(x)] with limits?

    I also have limits - zero and Pi.
    I seem to have obtained an expression for the integral but substituting Pi makes it troublesome.
    so, (apologies for the lack of LaTeX)
    let u = tan(x/2)

    u = sin(x/2) / cos(x/2)
    u cos(x/2) = sin(x/2)
    u^2 cos^2(x/2) = sin^2(x/2)
    u^2 cos^2(x/2) = (1 - cos^2(x/2)
    cos^2(x/2)(1 + u^2) = 1
    cos^2(x/2) = 1/ (1 + u^2)
    sin^2(x/2) = 1 - 1 /(1 + u^2) = u^2/(1 + u^2)
    cos x = cos^2(x/2) - sin^2(x/2)
    = 1 / (1 + u^2) - u^2/(1 + u^2) = (1 - u^2)/(1 + u^2)
    cos x = (1 - u^2)/(1 + u^2)
    when u = tan(x/2)
    x/2 = arctan(u)
    dx = 2du / (1 + u^2)

    substituting these into integral

    ∫ dx /(5 + 4 cos x)

    = ∫ 2du /(1 + u^2)(5 + 4(1 - u^2)/(1 + u^2)

    = ∫ 2du/ (5 + 5u^2 + 4 - 4u^2)

    = ∫ 2du/ (9 + u^2)

    = 2/9 ∫du/ [1 + (u/3)^2)]

    = (2/3) ∫(du/3)/ [1 + (u/3)^2)]

    = (2/3) arctan(u/3) + C

    substitute u/3 = (1/3)tan(x/2)

    = (2/3) arctan[1/3tan(x/2) ] + C

    but putting Pi into tan(x/2) makes the expression undefined! where did i go wrong?

    thank you
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Let u=\tan(\frac{x}{2}), then du=\frac{1}{2}\sec^2(\frac{x}{2})dx

    Now use the substitutions \sin(x)=\frac{2u}{u^2+1}, \cos(x)=\frac{1-u^2}{u^2+1}, and dx=\frac{2du}{u^2+1}. You'll find that some really good simplification happens.
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  3. #3
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    i did exactly that, please read my post
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  4. #4
    No one in Particular VonNemo19's Avatar
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    It's too hard to read. Check out this link: Learn Latex so we don't go crazy looking at that stuff.

    So your integrand Looks like this: \frac{2}{(u^2+1)\left[\frac{4(1-u^2)}{u^2+1}+4\right]}
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  5. #5
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    Hello, cassius1


    Your work is correct . . .


    Note that: . \tan\frac{\pi}{2} \:=\:\infty

    So we may write: . \arctan(\infty) \:=\:\frac{\pi}{2}


    We have: . \displaystyle \int^{\pi}_0 \frac{dx}{5 + 4\cos x} \;=\;\tfrac{2}{3}\arctan\left(\tfrac{1}{3}\tan\tfr  ac{x}{2}\right)\,\bigg]^{\pi}_0


    . . . . . . =\;\frac{2}{3}\arctan\left(\frac{1}{3}\tan\frac{\p  i}{2}\right) - \frac{2}{3}\arctan\left(\frac{1}{3}\tan 0\right)

    . . . . . . =\;\frac{2}{3}\underbrace{\arctan(\infty)} - \frac{2}{3}\arctan(0)

    . . . . . . =\quad\;\,\dfrac{2}{3}\left(\dfrac{\pi}{2}\right) \quad-\quad 0

    . . . . . . =\qquad\;\;\dfrac{\pi}{3}

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