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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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Evaluate the integral below by reversing the order of integration.

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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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