$ \huge \int f(g(x))\frac{\mathrm{d}g}{\mathrm{d}x}\mathrm{d}x \\ $ $\huge\begin{align*} u &= g(x) \\ \frac{\mathrm{d}u}{\mathrm{d}x} &= \frac{\mathrm{d}g}{{\mathrm{d}x}} \\ \int f(g(x))\frac{\mathrm{d}g}{\mathrm{d}x}\mathrm{d}x &= \int f(u)\mathrm{du} \\ \int f(u)\mathrm{du} &= F(g(x)) +C \end{align*}$ The process of converting an unknown integral of $f(x)$ to a known function $g(x)$. >[!example] >$\huge \begin{align} \int x\sin(x^2) \d x \end{align} $ >$\huge \begin{align} \let u &= x^2 \\ \let \d u &= 2x \\ \int x\sin \pa{x^2} \d x &= \int \frac{\d{u}}{2\d{x}} \sin(u) \d x\\ &= \frac{1}{2} \int \frac{\d{u}}{\d{x}} \sin(u) \d x\\ &=\frac{1}{2}\int\sin(u)\d u \\ &= \frac{1}{2} \cos(u) + C \\ &= \frac{1}{2} \cos\pa{x^2} \end{align} $