Substitution, also referred to as \(u\)-sub, is a technique of integration useful when the integrand is a composite of functions. Essentially it helps us infer the antiderivative of a function by “undoing” the chain rule of differentiation:
\(\displaystyle \int {\color{DarkGreen}{g'(x)}} f'\bigl({\color{DarkGreen}{g(x)}}\bigr)\,\mathrm{d}x = f\bigl({\color{DarkGreen}{g(x)}}\bigr) + C \)
Examples
\(\displaystyle \int 3x^2 \cos\bigl(x^3\bigr) \,\mathrm{d}x
\;\;=\;\; \int \cos\bigl({\color{DarkGreen}{x^3}}\bigr) {\color{DarkGreen}{3x^2}} \,\mathrm{d}x
\;\;=\;\; \int \cos\bigl({\color{DarkGreen}{u}}\bigr) {\color{DarkGreen}{\frac{\mathrm{d}u}{\cancel{\mathrm{d}x}}}} \,\cancel{\mathrm{d}x}
\;\;=\;\; \int \cos\bigl({\color{DarkGreen}{u}}\bigr) \,{\color{DarkGreen}{\mathrm{d}u}}
\;\;=\;\; \sin\bigl({\color{DarkGreen}{u}}\bigr) + C
\;\;=\;\; \sin\bigl(x^3\bigr) + C \)
\(\displaystyle \int x^4\sqrt{x^5+7} \,\mathrm{d}x
\;\;=\;\; \int \sqrt{{\color{DarkGreen}{x^5+7}}} \,\,{\color{DarkGreen}{x^4}} \,\mathrm{d}x
\;\;=\;\; \int \sqrt{{\color{DarkGreen}{u}}} \,\, {\color{DarkGreen}{\tfrac{1}{5}\frac{\mathrm{d}u}{\cancel{\mathrm{d}x}}}} \,\cancel{\mathrm{d}x}
\;\;=\;\; {\color{DarkGreen}{\tfrac{1}{5}}} \int \sqrt{{\color{DarkGreen}{u}}} \,{\color{DarkGreen}{\mathrm{d}u}}
\;\;=\;\; {\color{DarkGreen}{\tfrac{1}{5}}} \tfrac{2}{3} {\color{DarkGreen}{u}}^{3/2} + C
\;\;=\;\; \tfrac{2}{15} \bigl(x^5+7\bigr)^{3/2} + C \)
For a definite integral, the bounds must change in accordance with the substitution. If the previous example were instead a definite integral \(\int_0^2,\) those \(x\)-bounds of \(0\) and \(2\) would have to change to the \(u\)-bounds \(\int_7^{39}\) using the conversion \(u = x^5+7.\)