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Math Help - Polar Integral

  1. #1
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    Polar Integral

    I am having problems with this and was hoping somebody could assist. The polar function to integrate is:

    cos^2 7t
    cos^2 7t = 1/2(1+cos 14t)
    1/4|(1+cos 14t)^2 dt
    =1/4|1+2cos 14t+cos^2 14t dt
    =1/4|1dt + 1/2|cos 14t dt + 1/8|1dt + 1/8|cos 28t dt
    After evaluating this I get :-
    3t/8+(8sin 14t+sin 28t)/224 +C.

    However the answer given is:-
    t/2+(sin 14t)/28 + C
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by p75213 View Post
    I am having problems with this and was hoping somebody could assist. The polar function to integrate is:

    cos^2 7t
    cos^2 7t = 1/2(1+cos 14t)
    1/4|(1+cos 14t)^2 dt
    =1/4|1+2cos 14t+cos^2 14t dt
    =1/4|1dt + 1/2|cos 14t dt + 1/8|1dt + 1/8|cos 28t dt
    After evaluating this I get :-
    3t/8+(8sin 14t+sin 28t)/224 +C.

    However the answer given is:-
    t/2+(sin 14t)/28 + C
    you want to integrate
    \int \cos ^2 7t \; dt

    or

    \int (\cos ^2 7t )^2 \; dt
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  3. #3
    MHF Contributor matheagle's Avatar
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    Quote Originally Posted by p75213 View Post
    I am having problems with this and was hoping somebody could assist. The polar function to integrate is:

    cos^2 7t
    cos^2 7t = 1/2(1+cos 14t)
    1/4|(1+cos 14t)^2 dt
    =1/4|1+2cos 14t+cos^2 14t dt
    =1/4|1dt + 1/2|cos 14t dt + 1/8|1dt + 1/8|cos 28t dt
    After evaluating this I get :-
    3t/8+(8sin 14t+sin 28t)/224 +C.

    However the answer given is:-
    t/2+(sin 14t)/28 + C
    As captain america pointed out, you squared it twice.
    Afler using the half angle formula, there is NO need to square the function again.
    Just integrate 1/2(1+cos 14t) and you will get t/2+(sin 14t)/28 + C
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