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\$$\\int_0^\\var{a}\\int_{\\var{k}y}^{\\simplify{{k}*{a}}}\\var{b}sin(\\var{c}x^2)dx\\,dy\$$

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The limits are four lines \$$x=\\var{k}y,\\,\\,x=\\simplify{{k}*{a}}\$$ ,  \$$y=0\$$ and  \$$y=\\var{a}\$$

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These limits form a triangle, so we must keep: \$$x=\\simplify{{k}*{a}},\\,\\,x=\\var{k}y\$$   and   \$$y=0\$$

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On the inner integral we will have \$$y=0\$$ and \$$y=\\frac{x}{\\var{k}}\$$

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On the outer integral we have \$$x=\\simplify{{k}*{a}}\$$ and to this we add \$$x=0\$$

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\$$\\int_0^\\simplify{{a}*{k}}\\int_0^{\\frac{x}{\\var{k}}}\\var{b}sin(\\var{c}x^2)dy\\,dx\$$

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Inner integral:    \$$\\int_0^{\\frac{x}{\\var{k}}}\\var{b}sin(\\var{c}x^2)dy\$$

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\$$=\\var{b}sin(\\var{c}x^2).y\\,\\big|_0^{\\frac{x}{\\var{k}}}\$$

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\$$=\\var{b}sin(\\var{c}x^2)*\\frac{x}{\\var{k}}-\\var{b}sin(\\var{c}x^2)*0\$$

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\$$=\\frac{\\var{b}}{\\var{k}}xsin(\\var{c}x^2)\$$

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Outer integral:    \$$\\int_0^\\simplify{{a}*{k}}\\frac{\\var{b}}{\\var{k}}xsin(\\var{c}x^2)dx\$$

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This requires a u-substitution. Let \$$u=\\var{c}x^2\$$       \$$\\frac{du}{dx}=\\simplify{2*{c}}x\$$

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\$$du=\\simplify{2*{c}}x\\,dx\$$

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\$$\\frac{1}{\\simplify{2*{c}}x}du=dx\$$

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\$$\\implies\\int\\frac{\\var{b}}{\\var{k}}xsin(u)\\frac{1}{\\simplify{2*{c}}x}du\$$

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\$$=\\int\\frac{\\var{b}}{\\simplify{{k}*2*{c}}}sin(u)du\$$

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\$$=-\\simplify{{b}/({k}*2*{c})}cos(u)\$$

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\$$=-\\simplify{{b}/({k}*2*{c})}cos(\\var{c}x^2)\\,\\big|_0^\\simplify{{a}*{k}}\$$

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\$$=-\\simplify{{b}/({k}*2*{c})}cos(\\simplify{{c}*{a}^2*{k}^2})+\\simplify{{b}/({k}*2*{c})}cos(0)\$$

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\$$=-\\simplify{{b}/({k}*2*{c})}cos(\\simplify{{c}*{a}^2*{k}^2})+\\simplify{{b}/({k}*2*{c})}\$$

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\$$\\int_0^\\var{a}\\int_{\\var{k}y}^{\\simplify{{k}*{a}}}\\var{b}sin(\\var{c}x^2)dx\\,dy\$$