Calculus I

Substitution Rule

By letting

and

we can make a composite integrand

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 $du = \displaystyle\frac{1}{\sqrt{1 - x^2}} dx$ .
$\begin{displaymath}\begin{array}{ll} \displaystyle\int \frac{1}{\sqrt{1 - x^2 +... ...^{\frac{1}{2}} + C = 2 \sqrt{1 + \sin^{-1}x} + C \end{array}\end{displaymath}$