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$\\int \\cos(bx) \\mathrm{d}x = \\frac{\\sin(bx+a)}{b}$
\n$\\int \\sin(cx) \\mathrm{d}x = \\frac{-\\cos(cx+a)}{c}$
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\nGive your answer accurate to 2 decimal place.
", "precisionMessage": "You have not given your answer to the correct precision.", "allowFractions": false, "variableReplacements": [], "precision": "2", "maxValue": "(({c1[0]}/2)*{limit1}^2+cos({p1[1]}{limit1})/{p1[1]})-(({c1[0]}/2)*{limit0}^2+cos({p1[1]}*{limit0})/{p1[1]})", "minValue": "(({c1[0]}/2)*{limit1}^2+cos({p1[1]}{limit1})/{p1[1]})-(({c1[0]}/2)*{limit0}^2+cos({p1[1]}*{limit0})/{p1[1]})", "variableReplacementStrategy": "originalfirst", "strictPrecision": true, "correctAnswerFraction": false, "showCorrectAnswer": true, "precisionPartialCredit": 0, "scripts": {}, "marks": 1, "type": "numberentry", "showPrecisionHint": false}], "statement": "Find the following:
\nIn each case, you may assume that the constant of integration is 0
\nRemember to put your argument in brackets!
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