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Math Help - Problem on integration!! need answer for this.. pl reply how to solve

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    Problem on integration!! need answer for this.. pl reply how to solve

    how to solve this?? ∫(cos4x-cos4α)/(cosx-cosα)dx. Also the limits are given to be 0 to Pi. Reply me the entire solution.. Its urgent pls
    Last edited by kamaksh_ice; July 19th 2007 at 04:20 AM. Reason: Need not only the answer but also the solution
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  2. #2
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    Quote Originally Posted by kamaksh_ice View Post
    how to solve this?? ∫(cos4x-cos4α)/(cosx-cosα)dx. Also the limits are given to be 0 to Pi. Reply me the entire solution.. Its urgent pls
    Is this
    \int_0^{\pi} \frac{cos^4(x) - cos^4(a)}{cos(x) - cos(a)} dx
    or
    \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx

    -Dan
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    Answer me pls

    It is
    \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by kamaksh_ice View Post
    It is
    \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx
    This problem boils down to a factoring problem:

    cos(4x) = 8cos^4(x) - 8cos^2(x) + 1
    cos(4a) = 8cos^4(a) - 8cos^2(a) + 1

    So
    cos(4x) - cos(4a) = 8[cos^4(x) - cos^4(a)] - [cos^2(x) - cos^2(a)]

    = 8[cos^2(x) - cos^2(a)][cos^2(x) + cos^2(a)]- 8[cos^2(x) - cos^2(a)]

    = 8(cos(x) - cos(a))[cos(x) + cos(a)][cos^2(x) + cos^2(a)]  - 8(cos(x) - cos(a))[cos(x) + cos(a)]

    So
    \int_0^{\pi} \frac{cos(4x) - cos(4a)}{cos(x) - cos(a)} dx  = \int_0^{\pi} (8[cos(x) + cos(a)][cos^2(x) + cos^2(a)] - 8[cos(x) + cos(a)]) dx

    = 8 \int_0^{\pi}(cos^3(x) + cos(a) \cdot cos^2(x) - cos(x) + (cos^2(a) - cos(a))) dx

    all of which are now (relatively) elementary integrals.

    -Dan
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