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Calculating the integral of a function of the form $\\frac{\\cos(x)}{\\sin(x)+a}$ using integration by substitution.
", "licence": "None specified"}, "statement": "Calculate \\[ \\simplify{int(cos(x)/(sin(x)+{a}),x)}\\]
", "advice": "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.
\nFirstly, we must make a change of variables from $x$ to $u$, where $u$ is equal to the 'inner' function $g(x)$.
\nSo, for \\[\\simplify{int(cos(x)/(sin(x)+{a}),x)}\\]
\nlet $\\color{red}{u=\\simplify{sin(x)+{a}}}.$
\nNow, we need to calculate the differential, $du$, where \\[ du = \\left(\\frac{du}{dx}\\right)dx. \\]
\nDifferentiating $u$ with respect to $x$:
\n\\[ \\frac{du}{dx}= \\simplify{cos(x)}.\\]
\nTherefore, \\[ \\color{blue}{du = \\simplify{cos(x)}\\, dx}.\\]
\nWe can now rewrite the original integral in terms of $u$:
\n\\[ \\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}.\\]
\n(Note: It is important to see that both the function we are integrating, and the variable we are integrating with respect to, has changed.)
\n\\[ \\simplify[fractionNumbers]{int(1/u,u) = ln(abs(u)) + c}.\\]
\nFinally, we must rewrite our solution back in terms of the original variable $x$:
\n\\[ \\simplify[fractionNumbers]{ln(abs(u)) + c = ln(abs(sin(x)+{a})) + c}.\\]
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