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Thread: integrate........

  1. #1
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    integrate........

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  2. #2
    Member eXist's Avatar
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    Quote Originally Posted by gaurav1292 View Post
    Full problem please? And what are we integrating in terms of? And is $\displaystyle \alpha$ a constant?
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  3. #3
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    yeah a is constant..........integrate in terms of dx
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  4. #4
    Member eXist's Avatar
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    I'd say split it up:

    $\displaystyle
    \int\frac{cos2x - cos2\alpha}{cosx - cos\alpha}\,dx =
    \int\frac{cos2x}{cosx - cos\alpha}\,dx - \int\frac{cos2\alpha}{cosx - cos\alpha}\,dx
    $

    Can you do it now?
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  5. #5
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    i would say expand out cos2x - cos2a using the identity cos2x = cos^2 x - sin^2 x.

    so on the top you would have cos^2 x - sin^2 x - cos^2 a + sin^2 a. group the cos terms and sin terms together and split up the fraction into:
    (cos^2 x - cos^2 a) / (cos x - cos a) - (sin^2 x - sin^2 a) / (cos x - cos a)

    factor cos^2 x - cos^2 a into (cosx + cosa)(cosx - cosa) so you can cancel out terms on the top and bottom so your first integral would become: integral of (cosx + cosa). now for the second integral, remember the identity sin^2 x + cos^2 x = 1. so add and subtract a cos^2 x term and add and subtract a cos^2 a term. on top you have: sin^2 x - sin^2 a + cos^2 x - cos^2 x + cos^2 a - cos^2 a. now group the terms like so: (sin^2 x + cos^2 x) - (sin^2 a + cos^2 a) - cos^2 x + cos^2 a. using the trig identity, you get 1 - 1 - cos^2 x + cos^2 a = -cos^2 x + cos^2 a. factor out a negative so that the negative sign bewtween the 2 integrals becomes a positive. you now have (cos^2 x - cos^2 a) / (cosx - cosa). same as before, factor out the top and cancel out common factors so you are left with the integral of (cosx + cosa).

    so if you add that to the previous integral, you get 2(integral of (cosx + cosa)) = 2(sinx + xcosa) = 2sinx + 2x(cos a) + C
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