周次: 6 日期: April 1, 2022 节次: 2
In this lecture, we continue presenting the integration of rational functions. After finishing this part, we are going to discuss some trigonometric integration techniques.
Outline
§ 8.3 Integration of Rational Functions
Method of Partial Fractions ( for $f(x)/g(x)$ Proper)
- Factorize the denominator $g(x)$, obtain all its linear and quadratic factors, as well as their powers. In general, the denominator has the factorization looks like
$$ \dfrac{f(x)}{g(x)} = \dfrac{f(x)}{(x-r_1)^{m_1}\cdots (x-r_k)^{m_k} (x^2+p_1 x + q_1)^{n_1}\cdots(x^2+p_l x + q_l)^{n_l}} $$
where all the linear and quadratic factors are distinct. And $x^2+p_i x +q_i$ should not be factorized anymore ( called irreducible ).
**Note The feasibility of this factorization is guaranteed by the Fundamental Theorem of Algebra.**
- For each linear factor $(x-r)^m$ of $g(x)$, the fraction $f(x)/g(x)$ should include the following partial fractions, with undetermined coefficients $A_k$.
$$ \dfrac{A_1}{(x-r)} + \dfrac{A_2}{(x-r)^2} + \cdots \dfrac{A_m}{(x-r)^m} $$
- For each quadratic factor $(x^2+px+q)^n$ of $g(x)$, the obtained partial fractions should include
$$ \dfrac{B_1 x + C_1}{x^2+px+q} + \dfrac{B_2 x + C_2}{(x^2+px+q)^2} +\cdots + \dfrac{B_n x + C_n}{(x^2+px+q)^n} $$
- Set the original fraction $f(x)/g(x)$ equal to the sum of all these partial fractions. Clear the resulting equation of fractions and arrange the terms in decreasing powers of $x$.
- Equate the coefficients of corresponding powers of $x$ and solve the resulting equations for the undetermined coefficients.
Example 1. Distinct Linear Factors
Evaluate
$$ \int \dfrac{x^2+4x+1}{(x-1)(x+1)(x+3)}dx $$
using partial fractions.
( Solution: $\frac34 \ln|x-1|+\frac12\ln |x+1| -\frac14 \ln |x+3| + C$ .)
Example 2. A Repeated Linear Factor
Evaluate
$$ \int \dfrac{6x+7}{(x+2)^2}dx $$
( Solution: $6\ln|x+2| + \frac{5}{x+2} + C$ )
Example 3. Integrating an Improper Fraction
Evaluate
$$ \int \dfrac{2x^3-4x^2-x-3}{x^2-2x-3}dx $$
( Solution: $x^2 + 2\ln |x+1| + 3\ln |x-3| + C$ )
Example 4. Integrating with an Irreducible Quadratic Factor
Evaluate
$$ \int \dfrac{-2x+4}{(x^2+1)(x-1)^2}dx $$
(Solution: $\ln(x^2+1) + \arctan x - 2\ln|x-1| - \frac1{x-1} + C$ )
Example 5. A Repeated Irreducible Quadratic Factor
Evaluate
$$ \int \dfrac{dx}{x(x^2+1)^2}. $$
( Solution: $\ln \dfrac{|x|}{\sqrt{x^2+1}} + \dfrac1{2(x^2+1)} + C.$ )
The Heaviside “Cover-up” Method for Linear Factors
For proper fraction $f(x)/g(x)$, if $g(x) = (x-r_1) (x-r_2) \cdots (x-r_n)$ is a product of $n$ distinct linear factors, each raised to the first power, there is a quick way to find the undetermined coefficients.
Learn Heaviside “Cover-up” Method by Example
Find $A$, $B$ and $C$ in the partial-fraction expansion
$$ \dfrac{x^2+1}{(x-1)(x-2)(x-3)}= \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{x-3}. $$
Example 7. Integrating with the Heaviside Method
Evaluate
$$ \int \dfrac{x+4}{x^3+3x^2-10 x}dx. $$
( Solution : $-\frac25 \ln |x| + \frac37 \ln |x-2| - \frac1{35} \ln |x+5| + C$. )
Other ways to Determine the Coefficients
Example 8. Using Differentiation
Find $A$, $B$ and $C$ in the equation
$$ \frac{x-1}{(x+1)^3}= \dfrac{A}{x+1} + \dfrac{B}{(x+1)^2} + \dfrac{C}{(x+1)^3}. $$
( Solution : $A=0$, $B=1$ , $C=-2$ )
Example 9. Assigning Numerical Values to $x$ (Equivalent to “Cover-up” Method)
Find $A$, $B$ and $C$ in
$$ \dfrac{x^2+1}{(x-1)(x-2)(x-3)} = \dfrac{A}{x-1} + \dfrac{B}{x-2} + \dfrac{C}{x-3} $$
( Solution : $A=1$, $B=-5$, $C=5$ )
§ 8.4 Trigonometric Integrals
In this section, we consider the integrals involving algebraic combinations of the six basic trigonometric functions. For example the integration $\int \sec^2(x) dx$ , $\int \sec^3(x) \cot(x) dx$ etc.
Products of Powers of Sines and Cosines
Consider the integrals of the form
$$ \int \sin^m(x)\cos^n(x) dx $$
where $m$ and $n$ are nonnegative integers.
Case 1. If $m$ is odd, set $m = 2k+1$
$$ \int \sin^m(x) \cos^n(x)dx = -\int \sin^{2k}(x) \cos^n(x) d(\cos x) $$
It suggests changing $\sin^2(x)$ by $1-\cos^2(x)$ then the integration is reduced to a polynomial in $\cos(x)$
$$ -\int \left(1-\cos^2 x\right)^k \cos^n (x) d(\cos x) $$
Case 2. If $n$ is odd, set $n=2l+1$
Then similarly, we can change
$$ \int \sin^m(x) \cos^n(x) dx = \int \sin^m (x) \cos^{2l}(x) d(\sin x) = \int \sin ^m(x) \left(1-\sin^2 x\right)^l d(\sin x) $$
Then integrated by the substitution $y=\sin(x)$.
Case 3. If both $m$ and $n$ are even
In this case we use the substitution
$$ \sin^2(x) = \dfrac{1-\cos 2x}{2},\quad \cos^2(x) = \dfrac{1+\cos 2x}{2} $$
to reduce the integrand to one in lower powers of $\cos 2x$.
Example 1. $m$ is odd
Evaluate
$$ \int \sin^3 x \cos^2 x dx. $$
(Solution : $\frac15 \cos^5 x - \frac13 \cos^3 x + C.$ )
Example 2. $m$ is even, $n$ is odd
Evaluate
$$ \int \sin^2 x \cos^5 x dx $$
Solution.
Example 3. $m$ and $n$ are both even
Evaluate
$$ \int \sin^2(x) \cos^4(x) dx $$
(Solution: $\frac1{16} \left(x - \frac14 \sin 4x + \frac13 \sin^3(2x)\right) + C$ )
Eliminating Square Roots
Sometimes we can use the identity
$$ \cos^2 x = \dfrac{1+\cos 2x}{2} $$
to eliminate a square root.
Example 4.
Evaluate
$$ \int_0^{\pi/4}\sqrt{1+\cos 4x }dx. $$
( Solution: $\frac{\sqrt2}{2}$ )
Integrals of Powers of $\tan x$ and $\sec x$
To integrate higher powers of $\tan x$ and $\sec x$, we use the identities $\tan^2 x = \sec^2 x -1$ and $\sec^2 x = \tan^2 +1$ to reduce the higher powers to lower powers.
Example 5.
Evaluate
$$ \int \tan^4 x dx. $$
( Solution: $\frac13 \tan^3 x - \tan x + x + C$ )
Example 6.
Evaluate
$$ \int \sec^3 x dx. $$
( Solution: $\frac12 \sec x \tan x + \frac12 \ln |\sec x + \tan x| + C.$ )
Products of Sines and Cosines
The integrals
$$ \int \sin mx \sin nx , dx, \quad \int \sin mx \cos nx , dx, \quad \text{and} \int \cos mx \cos nx, dx $$
can be evaluated by integration by parts. But also more conveniently by using the following identities
$$ \sin mx \sin nx = - \frac12 \left[\cos (m+n) x -\cos (m-n) x\right] $$
$$ \sin mx \cos nx = \frac12 \left[\sin(m-n) x + \sin (m+n) x\right] $$
$$ \cos mx \cos nx = \frac12 \left[\cos (m-n) x + \cos (m+n) x\right] $$
Example 7.
Evaluate
$$ \int \sin 3x \cos 5x, dx $$
(Solution: $-\frac{\cos 8x}{16} + \frac{\cos 2x }{4} +C$ )