February 2010

Problem
Evaluate $\displaystyle \int \dfrac{(8x + 1) \, dx}{\sqrt{4x - 3}}$
 

In algebraic substitution we replace the variable of integration by a function of a new variable. A change in the variable on integration often reduces an integrand to an easier integrable form.
 

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

where $u = g(x)$
 

When $u$ and $v$ are differentiable functions of $x$, $d(uv) = u \, dv + v \, du\,$ or $\,u \, dv = d(uv) - v \, du$. When this is integrated we have
 

Integration by Parts
Integration by Substitution
Integration of Rational Fractions
Change of Limits with Change of Variable
 

In applying the formula (Example: Formula 1 below), it is important to note that the numerator du is the differential of the variable quantity u which appears squared inside the square root symbol. We mentally put the quantity under the radical into the form of the square of the constant minus the square of the variable.
 
1. $\displaystyle \int \dfrac{du}{\sqrt{a^2 - u^2}} = \arcsin \, \dfrac{u}{a} + C, \,\,\, a > 0$

2. $\displaystyle \int \dfrac{du}{a^2 + u^2} = \dfrac{1}{a}\arctan \, \dfrac{u}{a} + C$

3. $\displaystyle \int \dfrac{du}{u\sqrt{u^2 - a^2}} = \dfrac{1}{a} {\rm arcsec} \, \dfrac{u}{a} + C$
 

Basic Formulas

1. $\displaystyle \int \sin u \, du = -\cos u + C$

2. $\displaystyle \int \cos u \, du = \sin u + C$

3. $\displaystyle \int \sec^2 u \, du = \tan u + C$

4. $\displaystyle \int \csc^2 u \, du = -\cot u + C$

5. $\displaystyle \int \sec u \, \tan u \, du = \sec u + C$

6. $\displaystyle \int \csc u \, \cot u \, du = -\csc u + C$
 

There are two basic formulas for the integration of exponential functions.

1. $\displaystyle \int a^u \, du = \dfrac{a^u}{\ln a} + C, \,\, a > 0, \,\, a \neq 1$

2. $\displaystyle \int e^u \, du = e^u + C$
 

The limitation of the Power Formula $\displaystyle \int u^n \, du = \dfrac{u^{n + 1}}{n + 1} + C$, is when $n = -1$; this makes the right side of the equation indeterminate. This is where the logarithmic function comes in, note that $\displaystyle \int u^{-1} \, du = \displaystyle \int \frac{du}{u}$, and we can recall that $d(\ln u) = \dfrac{du}{u}$. Thus,
 

The General Power Formula as shown in Chapter 1 is in the form