Integration Techniques

Theorem 5.2.1 (Substitution)  
Let $f$ and $g$ be integrable functions and let $g$ also be a differentiable function. Then

$$\int f(g(x))g'(x)dx = \int f(u)du$$ (5.2.1)

with $u=g(x)$.

Proof.
Define

$$F(x):=\int f(x) dx$$ (5.2.2)

or in other terms: $F'=f$. So

$$F(g(x)):=\int f(g(x)) dx$$ (5.2.3)

Applying the chain rule and the fact that $F'=f$ gives us

$$\frac{d{}}{dx} F(g(x)) = F'(g(x)) \cdot g'(x) = f(g(x)) \cdot g'(x)$$ (5.2.4)

Putting $u=g(x)$ and then integrating the left and the right hand side then gives

$$\frac{d{}}{dx} F(u) = F(g(x)) = \int f(g(x)) \cdot g'(x) dx$$ (5.2.5)

$\qedsymbol$

Remark 5.2.2  
Formula (5.2.1) can be more than just a bit confusing. When you use substitution, you essentially change the variable you are integrating over. Instead of integrating over $x$, you are now integrating over $g(x)$, so you also have to adapt the integration boundaries when calculating a definite integral:

$$\int_a^b f(g(x))g'(x)dx = \int_{g(a)}^{g(b)} f(u)du$$ (5.2.6)

Theorem 5.2.3 (Integration by Parts)  
Let $f$ and $g$ be differentiable functions. Then

$$\int f(x)g'(x)dx = f(x)g(x) - \int f'(x)g(x)dx$$ (5.2.7)

Proof.
Remember the product rule:

$$\frac{d{}}{dx}[f(x)g(x)] = f'(x)g(x) + f(x)g'(x)$$ (5.2.8)

Integrating the left and the right side gives the desired result. $\qedsymbol$

Remark 5.2.4  
Again, the comment has to be made that for definite integral, one should not forget to put in the correct boundaries:

$$\int_a^b f(x)g'(x)dx = f(b)g(b) - f(a)g(a) - \int_a^b f'(x)g(x)dx$$ (5.2.9)

Remark 5.2.5  
And this concludes pretty much all the basic techniques that you have at hand to calculate an integral. Apart from substitution and integration by parts, there are however a lot of standard tricks which are not really within the realm of integration theory, but still can make your integrating life a lot easier. Here we present just a few of them:

• Look for obvious solutions. Maybe you are being asked to integrate an odd function (a function for which $f(-x)=-f(x)$) over a symmetric interval (an interval like $[-x,x]$). Then the picture is something like
  200pt
  Look e.g at the integral

  $$\int_{-a}^a e^{-x^2} \sin(x)dx = 0$$ (5.2.10)

  
  
  for any $a$ we choose. Remarking that the positive and the negative parts cancel out saves us having to calculate a rather unpleasant integral.
• Use elementary algebra to rewrite the integrand in a simpler form, e.g.

  $$\int \sqrt{x} ( 1 + \sqrt{x} ) dx = \int (\sqrt{x} + x) dx = \frac{2}{3}x\sqrt{x} + \frac{1}{2} x^2$$ (5.2.11)

  
  

• Sometimes doing a long division before integrating simplifies things a little. The integral

  $$\int \frac{x^3}{x+1} dx$$ (5.2.12)

  
  
  looks quite unpleasant because no simplifying is possible. Rational functionals are only easy to deal with if the degree of the numerator happens to be one less than the degree of the denominator, and we just get a logarithm. In this case, doing the long division of $x^3$ by $x-1$ gives

  $$x^3 = x^2-x+1-\frac{1}{x+1}$$ (5.2.13)

  
  
  and therefore
  

  $$\int x^3 dx = \int x^2-x+1 dx - \int\frac{1}{x+1}dx \cr$$ (5.2.14)

  
  

• Use one of the 4 million trig formulas you were pestered with in your more juvenile days to rewrite the integrand in a simpler form, e.g.

  $$\begin{aligned}\int \sin(x) \cos(x) dx &= \frac{1}{2} \int \sin... ...dy \text{ with $y=2x$} \\ &= - \frac{1}{4} \cos(2x) \end{aligned}$$

  
  
  The most useful ones tend to be

  $$\sin^2(\theta)=\frac{1-\cos(2\theta)}{2} \hbox{\ \ and\ \ } \cos^2(\theta)=\frac{1+\cos(2\theta)}{2}$$ (5.2.16)

  
  
  as the following example will illustrate.
• A standard substitution technique used to integrate root functions goes as follows:
  • In $\sqrt{a^2-x^2}$, substitute $x=a\sin(\theta)$ to obtain $\sqrt{a^2-x^2}=\sqrt{a^2-a^2\cos^2(\theta)}=a\cos(\theta)$.
  • In $\sqrt{a^2+x^2}$, substitute $x=a\tan(\theta)$ to obtain $\sqrt{a^2+x^2}=\sqrt{a^2+a^2\tan^2(\theta)}=a\sec(\theta)$.
  • In $\sqrt{x^2-a^2}$, substitute $x=a\sec(\theta)$ to obtain $\sqrt{x^2-a^2}=\sqrt{a^2\sec^2(\theta)-a^2}=a\tan(\theta)$.
  Example:
  

  $$\int \sqrt{4-x^2} dx = \int \sqrt{4-4\sin^2(\theta)} (2\cos(\theta)d\theta) \cr$$ (5.2.17)

  
  
  Using our substitution $x=2\sin(\theta)$ and therefore $\theta=\arcsin(\frac{x}{2})$. Furthermore, we know

  $$\sin(\theta)=\frac{x}{2} \hbox{\ \ and\ \ }\cos(\theta)=\sqrt{1-\frac{x^2}{4}}$$ (5.2.18)

  
  
  which gives
  

  $$\int \sqrt{4-x^2} dx = 2\arcsin\left(\frac{x}{2}\right) + 2 \cdot \frac{x}{2} \cdot \sqrt{1-\frac{x^2}{4}} \cr$$ (5.2.19)

  
  

• What is strictly speaking also part of ``elementary algebra" is the use of partial fractions. They are very useful when having to calculate the integral of a rational function:

  $$\int \frac{1}{x^2 - 1} dx$$ (5.2.20)

  
  
  Would be an easy enough integral if only there where an $x$ in the numerator, but now it looks kind of difficult. The standard way to tackle this problem is to split up the fraction in different ones:

  $$\frac{1}{x^2-1} = \frac{A}{x+1} + \frac{B}{x-1}$$ (5.2.21)

  
  
  This means that

  $$\frac{1}{x^2-1} = \frac{Ax-A+Bx+B}{x^2-1}$$ (5.2.22)

  
  
  So we then have to solve the system of equations for $A$ and $B$:

  $$\begin{cases}A+B = 0 \\ -A+B = 1 \end{cases}$$ (5.2.23)

  
  
  The solution of this system is $A=-\frac{1}{2}$ and $B=\frac{1}{2}$ so

  $$\frac{1}{x^2-1} = \frac{-1}{2(x+1)} + \frac{1}{2(x-1)}$$ (5.2.24)

  
  
  and therefore

  $$\begin{aligned}\int\frac{1}{x^2-1} &= -\int\frac{1}{2(x+1)} + \... ...n \vert 2x+2\vert + \frac{1}{2} \ln \vert 2x-2\vert \end{aligned}$$

  
  
  There are a lot of much more complicated integrals that can be solved using this method, but this calculation gives the general idea of it.

• When everything else fails, there is always an integral table ...

Table of Integration Formulas

$$\int x^n dx = \frac{x^{n+1}}{n+1} \text{for $n\not=1$}$$	$$\int \frac{1}{x}dx = \ln \vert x\vert$$
$$\int e^x dx = e^x$$	$$\int a^x dx = \frac{a^x}{\ln a}$$
$$\int \sin(x) dx = - \cos(x)$$	$$\int \cos(x) dx = \sin(x)$$
$$\int \sec^2(x) dx = \tan(x)$$	$$\int \csc^2(x) dx = -\cot(x)$$
$$\int \sec(x) \tan(x) dx = \sec(x)$$	$$\int \csc(x) \cot(x) dx = -\csc(x)$$
$$\int \sec(x) dx = \ln \vert\sec(x) + \tan(x)\vert$$	$$\int \csc(x) dx = \ln \vert\csc(x) - \cot(x)\vert$$
$$\int \tan(x) dx = \ln \vert\sec(x)\vert$$	$$\int \cot(x) dx = \ln \vert\sin(x)\vert$$
$$\int \sinh(x) dx = \cosh(x)$$	$$\int \cosh(x) dx = \sinh(x)$$
$$\int \frac{dx}{x^2+a^2} = \frac{1}{a}\arctan\left(\frac{x}{a}\right)$$	$$\int \frac{dx}{\sqrt{a^2-x^2}} = \arcsin\left(\frac{x}{a}\right)$$
$$\int \frac{dx}{x^2-a^2} = \frac{1}{2a}\ln\left\vert\frac{x-a}{x+a}\right\vert$$	$$\int \frac{dx}{\sqrt{a^2 \pm x^2}} = \ln \vert x+\sqrt{x^2 \pm a^2} \vert$$