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Calculating the definite integral $\\int_{n_1}^{n_2}a \\sin(bx) dx$, where $n_1$ and $n_2$ are multiples of $\\frac{\\pi}{12}$.

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Evaluate \\[ \\int_\\simplify{{n_1}pi/12}^\\simplify{{n_2}pi/12}\\simplify{{a}sin({b}x)}dx\\]

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Integrating a function of the form  \\[ f(x)=\\sin(kx) \\] has the integral \\[ \\int_a^b \\sin(kx) dx  =  \\left[-\\frac{1}{k}\\cos(kx)\\right]_a^b,\\]

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and \\[\\int_a^b kf(x) dx = k \\int_a^b f(x) dx.\\]

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Therefore,

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\\[ \\begin{split} \\simplify{defint({a}sin({b}x),x,{n_1}pi/12,{n_2}pi/12)} &\\,= \\simplify{{a}defint(sin({b}x),x,{n_1}pi/12,{n_2}pi/12)}, \\\\ &\\,= \\left[ \\simplify{-{a}/{b}cos({b}x)}\\right]_\\simplify{{n_1}pi/12}^\\simplify{{n_2}pi/12}, \\\\ &\\,= \\left[\\simplify[all,!trig,fractionNumbers]{-{a/b}cos({simp2}pi)}\\right]-\\left[\\simplify[all,!trig,fractionNumbers]{-{a/b} cos({simp1}pi)}\\right] \\\\ &\\,=\\left[\\simplify[all,fractionNumbers]{-{a*frac2/b}}\\right]-\\left[\\simplify[all,fractionNumbers]{-{a*frac1/b}}\\right]\\\\ &\\,=\\simplify[all,fractionNumbers]{{(-a*frac2+a*frac1)/b}} \\end{split} \\]

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[[0]] (Give your answer as a fraction or to 2 decimal places, where necessary)

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