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$\\int \\cos(bx) \\mathrm{d}x = \\frac{\\sin(bx+a)}{b}$

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$\\int \\sin(cx) \\mathrm{d}x = \\frac{-\\cos(cx+a)}{c}$

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Evaluate: $\\int_{\\var{limit0}}^{\\var{limit1}} (\\simplify{{c1[0]}x - sin({p1[1]}x)} \\mathrm){d}x$. (the angle in the trig. is in radians.)

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Give your answer accurate to 2 decimal place.

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Find the following:

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In each case, you may assume that the constant of integration is 0

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Remember to put your argument in brackets!

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