Calculus I

Substitution Rule

By letting ${u = g(x)}$

and ${}du = g'(x)\,dx$ we can make a composite integrand ${}f(g(x)) g'(x)$

into

by the following substitution rule.

$\displaystyle {\int f(g(x)) g'(x)\,dx = \int f(u)\,du}$

Let ${}u = \sqrt{x}$ and ${}du = \frac{1}{2\sqrt{x}}\,dx$ .
${}\displaystyle\int\frac{\sin\sqrt{x}}{\sqrt{x}}\,dx = 2\int \sin u\,du = -2 \cos u + C = -2 \cos\sqrt{x} + C$
Let ${}u = \ln x$ and ${}du = \displaystyle\frac{1}{x}\,dx$ .
${}\displaystyle\int \frac{\cos(\ln x)}{x}\,dx = \int \cos u\,du = \sin u + C = \sin(\ln x) + C$
Let ${}u = 1 + \sin^{-1}x$ and ${}du = \displaystyle\frac{1}{\sqrt{1 - x^2}}\,dx$ .
$\begin{displaymath}\begin{array}{ll} \displaystyle\int \frac{1}{\sqrt{1 - x^2 +... ...^{\frac{1}{2}} + C = 2 \sqrt{1 + \sin^{-1}x} + C \end{array}\end{displaymath}$