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Evaluate the following definite integral correct to three decimal places:

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\$$\\int_\\var{a}^\\var{b}\\var{m}cos^4(\\var{k}t)dt\$$

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\$$Answer =\$$ [[0]]

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\$$\\int_\\var{a}^\\var{b}\\var{m}cos^4(\\var{k}t)dt\$$

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\$$=\\int_\\var{a}^\\var{b}\\var{m}cos^2(\\var{k}t)cos^2(\\var{k}t)dt\$$

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\$$=\\int_\\var{a}^\\var{b}\\var{m}\\left(\\frac{1}{2}(1+cos(\\simplify{2*{k}}t))\\right)\\left(\\frac{1}{2}(1+cos(\\simplify{2*{k}}t))\\right)\$$

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\$$=\\int_\\var{a}^\\var{b}\\frac{\\var{m}}{4}\\left(1+cos(\\simplify{2*{k}}t)+cos(\\simplify{2*{k}}t)+cos^2(\\simplify{2*{k}}t)\\right)\$$

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\$$=\\int_\\var{a}^\\var{b}\\frac{\\var{m}}{4}\\left(1+2cos(\\simplify{2*{k}}t)+\\frac{1}{2}(1+cos(\\simplify{4*{k}}t))\\right)\$$

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\$$=\\int_\\var{a}^\\var{b}\\frac{\\var{m}}{4}\\left(\\frac{3}{2}+2cos(\\simplify{2*{k}}t)+\\frac{1}{2}cos(\\simplify{4*{k}}t)\\right)\$$

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\$$=\\int_\\var{a}^\\var{b}\\frac{\\simplify{3*{m}}}{8}+\\frac{\\var{m}}{2}cos(\\simplify{2*{k}}t)+\\frac{\\var{m}}{8}cos(\\simplify{4*{k}}t)\$$

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\$$=\\frac{\\simplify{3*{m}}}{8}t+\\frac{\\var{m}}{\\simplify{4*{k}}}sin(\\simplify{2*{k}}t)+\\frac{\\var{m}}{\\simplify{32*{k}}}sin(\\simplify{4*{k}}t)\\Big|_\\var{a}^\\var{b}\$$

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\$$=\\frac{\\simplify{3*{m}*{b}}}{8}+\\frac{\\var{m}}{\\simplify{4*{k}}}sin(\\simplify{2*{k}*{b}})+\\frac{\\var{m}}{\\simplify{32*{k}}}sin(\\simplify{4*{k}*{b}})-\\frac{\\simplify{3*{m}*{a}}}{8}-\\frac{\\var{m}}{\\simplify{4*{k}}}sin(\\simplify{2*{k}*{a}})-\\frac{\\var{m}}{\\simplify{32*{k}}}sin(\\simplify{4*{k}*{a}})\$$

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