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Calculating the integral of a function of the form $\\frac{\\cos(x)}{\\sin(x)+a}$ using integration by substitution.

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Calculate \\[ \\simplify{int(cos(x)/(sin(x)+{a}),x)}\\]

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Since this integral is of the form \\[ \\int g'(x)f(g(x))\\,dx,\\] we can use the method of substitution to calculate the solution. 

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Firstly, we must make a change of variables from $x$ to $u$, where $u$ is equal to the 'inner' function $g(x)$.

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So, for \\[\\simplify{int(cos(x)/(sin(x)+{a}),x)}\\]

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let $\\color{red}{u=\\simplify{sin(x)+{a}}}.$

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Now, we need to calculate the differential, $du$, where \\[ du = \\left(\\frac{du}{dx}\\right)dx. \\]

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Differentiating $u$ with respect to $x$:

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\\[ \\frac{du}{dx}= \\simplify{cos(x)}.\\]

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Therefore, \\[ \\color{blue}{du = \\simplify{cos(x)}\\, dx}.\\]

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We can now rewrite the original integral in terms of $u$:

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\\[ \\int \\frac{\\color{blue}{\\cos(x)}}{\\color{red}{\\simplify{sin(x)+{a}}}}\\color{blue}{\\text{d}x} = \\int \\frac{1}{\\color{red}{u}}\\color{blue}{\\text{d}u}.\\]

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(Note: It is important to see that both the function we are integrating, and the variable we are integrating with respect to, has changed.)

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\\[ \\simplify[fractionNumbers]{int(1/u,u) = ln(abs(u)) + c}.\\]

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Finally, we must rewrite our solution back in terms of the original variable $x$:

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\\[ \\simplify[fractionNumbers]{ln(abs(u)) + c = ln(abs(sin(x)+{a})) + c}.\\]

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