Integration Techniques

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Integration Techniques

No silver bullet. Here is the review of basic integration techniques, together with some of the challenging problems presented at University of North Texas Integration Bee. You will see that even computer algebra systems such as Maple and Mathematica are not tough enough to complete them. Nevertheless, you could outperform smart software with the basic techniques we have learned.

Derivatives and integrals of basic functions.

Functions Derivatives Indefinite integrals
Fundamental theorem

of calculus

   
  $\displaystyle\frac{d}{dx} F(x) = f(x)$ $\displaystyle\int f(x) dx = F(x) + C$
Power and logarithmic

functions

   
  $\displaystyle\frac{d}{dx} x^n = n x^{n-1}$ $\displaystyle\int x^n dx = \frac{x^{n+1}}{n+1} + C$ ($n \neq -1$)
  $\displaystyle\frac{d}{dx} \ln \vert x\vert = \frac{1}{x}$ $\displaystyle\int \frac{1}{x} dx = \ln \vert x\vert + C$
Exponential functions    
  $\displaystyle\frac{d}{dx} e^x = e^x$ $\displaystyle\int e^x dx = e^x + C$
Trigonometric functions    
  $\displaystyle\frac{d}{dx} \sin x = \cos x$ $\displaystyle\int \cos x dx = \sin x + C$
  $\displaystyle\frac{d}{dx} \cos x = -\sin x$ $\displaystyle\int \sin x dx = -\cos x + C$
  $\displaystyle\frac{d}{dx} \tan x = \frac{1}{\cos^2 x} = \sec^2 x$ $\displaystyle\int \sec^2 x\,dx =
\int \frac{1}{\cos^2 x}\,dx = \tan x + C$
  $\displaystyle\frac{d}{dx} \cot x = -\frac{1}{\sin^2 x} = -\csc^2 x$ $\displaystyle\int \csc^2 x\,dx =
\int \frac{1}{\sin^2 x}\,dx = -\cot x + C$
  $\displaystyle\frac{d}{dx} \sec x = \sec x \tan x$ $\displaystyle\int \sec x \tan x\,dx = \sec x + C$
Hyperbolic functions    
  $\displaystyle\frac{d}{dx} \sinh x = \cosh x$ $\displaystyle\int \cosh x dx = \sinh x + C$
  $\displaystyle\frac{d}{dx} \cosh x = \sinh x$ $\displaystyle\int \sinh x dx = \cosh x + C$
  $\displaystyle\frac{d}{dx} \tanh x = \frac{1}{\cosh^2 x}$ $\displaystyle\int \frac{1}{\cosh^2 x} dx = \tanh x + C$
  $\displaystyle\frac{d}{dx} \coth x = -\frac{1}{\sinh^2 x}$ $\displaystyle\int \frac{1}{\sinh^2 x} dx = -\coth x + C$
Inverse trigonometric

functions

   
  $\displaystyle\frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}}$ $\displaystyle\int \frac{1}{\sqrt{1 - x^2}} dx = \sin^{-1} x + C$
  $\displaystyle\frac{d}{dx} \cos^{-1} x = -\frac{1}{\sqrt{1 - x^2}}$  
  $\displaystyle\frac{d}{dx} \tan^{-1} x = \frac{1}{1 + x^2}$ $\displaystyle\int \frac{1}{1 + x^2} dx = \tan^{-1} x + C$
Inverse hyperbolic

functions

   
  $\displaystyle\frac{d}{dx} \sinh^{-1} x = \frac{1}{\sqrt{1 + x^2}}$ $\displaystyle\int \frac{1}{\sqrt{1 + x^2}} dx = \sinh^{-1} x + C$
  $\displaystyle\frac{d}{dx} \cosh^{-1} x = \frac{1}{\sqrt{x^2 - 1}}$ $\displaystyle\int \frac{1}{\sqrt{x^2 - 1}} dx = \cosh^{-1} x + C$
  $\displaystyle\frac{d}{dx} \tanh^{-1} x = \frac{1}{1 - x^2}$ $\displaystyle\int \frac{1}{1 - x^2} dx = \tanh^{-1} x + C$

Formulas and identities of basic functions.

Logarithmic and

Exponential functions

   
  $\displaystyle\log_{a} x = \frac{\ln x}{\ln a}$ $\displaystyle a^x = e^{x\ln a}$
Trigonometric functions    
  $\displaystyle\sec x = \frac{1}{\cos x}$ $\displaystyle\csc x = \frac{1}{\sin x}$
  $\displaystyle\tan x = \frac{\sin x}{\cos x}$ $\displaystyle\cot x = \frac{\cos x}{\sin x}$
  $\displaystyle\cos^2 x + \sin^2 x = 1$ $\displaystyle 1 - \cos^2 x = \sin^2 x$
  $\displaystyle 1 + \tan^2 x = \sec^2 x$ $\displaystyle \sec^2 - 1 = \tan^2 x$
  $\displaystyle \sin 2x = 2 \sin x \cos x$ $\displaystyle\sin x = 2\sin\frac{x}{2} \cos\frac{x}{2}
= \frac{2\tan\frac{x}{2}}{1 + \tan^2\frac{x}{2}}^\dagger$
  $\displaystyle \cos 2x = \cos^2 x - \sin^2 x$ $\displaystyle\cos x = \cos^2\frac{x}{2} - \sin^2\frac{x}{2}
= \frac{1 - \tan^2\frac{x}{2}}{1 + \tan^2\frac{x}{2}}^\dagger$
  $\displaystyle\sin^2 x = \frac{1 - \cos 2x}{2}$ $\displaystyle\cos^2 x = \frac{1 + \cos 2x}{2}$
  $\displaystyle A \cos x + B \sin x = \sqrt{A^2 + B^2}\cos(x - \theta)$ where $A > 0$ and $\theta = \tan^{-1}(B/A)$.
Hyperbolic functions    
  $\displaystyle\sinh x = \frac{e^x - e^{-x}}{2}$ $\displaystyle\sinh^{-1} x = \ln\left\vert x + \sqrt{x^2 + 1}\right\vert$
  $\displaystyle\cosh x = \frac{e^x + e^{-x}}{2}$ $\displaystyle\cosh^{-1} x = \ln\left\vert x + \sqrt{x^2 - 1}\right\vert$
  $\displaystyle\tanh x
= \frac{\sinh x}{\cosh x} = \frac{e^x - e^{-x}}{e^x + e^{-x}}$ $\displaystyle\tanh^{-1} x = \frac{1}{2}\ln\left\vert\frac{1 + x}{1 - x}\right\vert$
  $\displaystyle\cosh^2 x - \sinh^2 x = 1$  



$\dagger$ You may find them useful in the following ``half-angle tangent'' substitution:


Substitution   Identities  
       
$t = \displaystyle\tan\left(\frac{x}{2}\right)$, $\displaystyle\frac{x}{2} = \tan^{-1} t$; $\sin x = \displaystyle\frac{2t}{1 + t^2}$, $\cos x = \displaystyle\frac{1 - t^2}{1 + t^2}$, $\displaystyle dx = \frac{2}{1 + t^2} dt$.


\begin{align*}
& \mbox{Example. }
\int \frac{1 + \sin x}{1 + \cos x} dx
= \int ...
...an\frac{x}{2} + \ln \left\vert 1 + \tan^2\frac{x}{2}\right\vert + C
\end{align*}

Strategy for integration with sample problems.

A. Substitution rules.

(A.1).
$\displaystyle\int y \frac{dt}{dx} dx = \int y dt$


\begin{align*}
& \mbox{(a) }
\int\frac{\sin\sqrt{x}}{\sqrt{x}} dx
= 2\int \sin ...
...ce{5.0in} \\
&\hspace{0.4in}
= -2 \cos t + C = -2 \cos\sqrt{x} + C
\end{align*}


\begin{align*}
& \mbox{(b) }
\int \frac{\cos(\ln x)}{x} dx
= \int \cos t dt
\h...
...$}\hspace{5.0in} \\
&\hspace{0.4in}
= \sin t + C = \sin(\ln x) + C
\end{align*}


\begin{align*}
& \mbox{(c) }
\int \frac{\sin x}{1 + \sin x} dx
= \int \frac{\si...
... \frac{1}{t} - [\tan x - x] + C
= \frac{1}{\cos x} - \tan x + x + C
\end{align*}


\begin{align*}
& \mbox{(d) }
\int \frac{1}{\sqrt{1 - x^2 + \sin^{-1}x - x^2\sin^...
...hspace{0.4in}
= 2 t^{\frac{1}{2}} + C
= 2 \sqrt{1 + \sin^{-1}x} + C
\end{align*}


\begin{align*}
& \mbox{(e) }
\int \frac{1}{\sin x + \cos x} dx
= \frac{1}{\sqrt...
...frac{1}{\sqrt{2}} \tanh^{-1} \sin\left(x - \frac{\pi}{4}\right) + C
\end{align*}

(A.2).
$\displaystyle\int y dx = \int y \frac{dx}{dt} dt$


\begin{align*}
& \mbox{(a) }
\int x^2\sqrt{x + 4} dx
= \int (t^2 - 4)^2 t 2t\...
... \frac{16}{5}(x+4)^\frac{5}{2}
+ \frac{32}{3}(x+4)^\frac{3}{2} + C
\end{align*}


\begin{align*}
& \mbox{(b) }
\int \sqrt{1 - e^x} dx
= \int t \frac{2t}{t^2 - 1...
...{-1} t\right] + C
= 2\sqrt{1 - e^x} - 2\tanh^{-1}\sqrt{1 - e^x} + C
\end{align*}


\begin{align*}
& \mbox{(c) }
\int \frac{1}{x^4 - 16} dx
= \frac{1}{8}\int \frac...
...c{x}{2}\right)
- \frac{1}{16}\tan^{-1}\left(\frac{x}{2}\right) + C
\end{align*}


\begin{align*}
& \mbox{(d) }
\int \frac{x^2}{\sqrt{x-1}} dx
= 2\int (t^2 + 1)^2...
...frac{5}{2} + \frac{4}{3}(x-1)^\frac{3}{2}
+ 2(x-1)^\frac{1}{2} + C
\end{align*}

(A.3).
$\displaystyle\int \frac{1}{y} \frac{dy}{dx} dx = \ln\vert y\vert + C$
\begin{align*}
& \mbox{(a) }
\int \sec x  dx
= \int \sec x  \frac{\sec x + \ta...
...e{0.4in}
= \ln \vert y\vert + C = \ln \vert\sec x + \tan x\vert + C
\end{align*}

(A.4).
Trigonometric substitution.


Substitution   Identities  
       
$x = a \sin t$, $\displaystyle t = \sin^{-1} \left(\frac{x}{a}\right)$; $a^2 - x^2 = a^2 \cos^2 t$, $dx = a \cos t  dt$.
$x = a \tan t$ $\displaystyle t = \tan^{-1} \left(\frac{x}{a}\right)$; $a^2 + x^2 = a^2 \sec^2 t$, $dx = a \sec^2 t  dt$.
$x = a \sec t$ $\displaystyle t = \cos^{-1} \left(\frac{a}{x}\right)$; $x^2 - a^2 = a^2 \tan^2 t$, $dx = a \sec t \tan t  dt$.


\begin{align*}
& \mbox{(a) }
\displaystyle\int \frac{1}{\sqrt{1 - x^2}} dx =
\i...
... C
= \ln\left\vert x + \sqrt{x^2 - 1}\right\vert
= \cosh^{-1} x + C
\end{align*}

B. Integration by parts: $\displaystyle\int u dv = uv - \int v du$


\begin{align*}
& \mbox{(a) }
\int \sqrt{x}\ln x dx
= \frac{2}{3} x^\frac{3}{2}...
...}
= \frac{2}{3} x^\frac{3}{2} \ln x - \frac{4}{9} x^\frac{3}{2} + C
\end{align*}


\begin{align*}
& \mbox{(b) }
\int e^x\sin x dx
= e^x \sin x - \int e^x\cos x d...
...displaystyle\int e^x\sin x dx = \frac{e^2}{2}(\sin x - \cos x)$. }
\end{align*}


\begin{align*}
& \mbox{(c) }
\int x^3 e^{-x^2} dx
= \frac{1}{2}\int t e^{-t} d...
...\right] + C
= - \frac{1}{2} x^2 e^{-x^2} - \frac{1}{2} e^{-x^2} + C
\end{align*}


\begin{align*}
& \mbox{(d) }
\int \frac{x}{2 + e^x + e^{-x}} dx
= 4 \int \frac{...
...\vert s\vert + C
= (x/2) \tanh (x/2) - \ln \vert\cosh(x/2)\vert + C
\end{align*}


\begin{align*}
& \mbox{(e) }
\int x^3\sin x^2 dx
= \frac{1}{2}\int t\sin t dt
...
...{1}{2}\sin t + C
= -\frac{x^2}{2}\cos x^2 + \frac{1}{2}\sin x^2 + C
\end{align*}


\begin{align*}
& \mbox{(f) }
\int \sin x \ln\vert\cos x\vert dx
= \int -\ln\ver...
... \ln\vert t\vert + t + C
= -\cos x \ln\vert\cos x\vert + \cos x + C
\end{align*}


\begin{align*}
& \mbox{(g) }
\int \frac{x^2}{(x^2 + 8)^\frac{3}{2}} dx
= -x(x^2...
...{x}{\sqrt{x^2 + 8}} + \sinh^{-1}\left(\frac{x}{\sqrt{8}}\right) + C
\end{align*}

C. Integration of rational functions.

(C.1).
Partial fractions

$\displaystyle\frac{1}{x^3 + 1}
= \frac{1}{3}\left[\frac{1}{x + 1} - \frac{x - 2}{x^2 - x+1}\right]$ ; $\displaystyle\frac{1}{x^3 - 1}
= \frac{1}{3}\left[\frac{1}{x - 1} - \frac{x + 2}{x^2 + x+1}\right]$ .
$\displaystyle\frac{1}{x^4 - 1}
= \frac{1}{2}\left[\frac{1}{x^2-1} - \frac{1}{x^2+1}\right]$ ; $\displaystyle\frac{1}{x^4 + 1}
= \frac{\sqrt{2}}{4}\left[
\frac{x+\sqrt{2}}{x^2+\sqrt{2}x+1} - \frac{x-\sqrt{2}}{x^2-\sqrt{2}x+1}
\right]^\star$ .


$\displaystyle\left[
^\star\mbox{ Use }\:
x^4 + 1 = (x^2+1)^2 - (\sqrt{2} x)^2
= (x^2+\sqrt{2}x+1)(x^2-\sqrt{2}x+1)
\:\right]$


\begin{align*}
&\mbox{(a) }
\int \frac{1}{x^3 + 8} dx
= \frac{1}{4}\int \frac{1...
... +\frac{\sqrt{3}}{12}\tan^{-1}\left(\frac{x-1}{\sqrt{3}}\right) + C
\end{align*}

(C.2).
$\displaystyle\int \frac{x+a}{x^2+bx+c} dx
= {\textstyle\left(\frac{1}{2}\right...
...c\vert^{\:*}
+ {\textstyle\left(a-\frac{b}{2}\right)}\int\frac{1}{x^2+bx+c} dx$


$\left[
^{*}\mbox{ Use }\:
\frac{x+a}{x^2+bx+c} =
{\textstyle\left(\frac{1}{2}\r...
...{ and (A.1) with }\:
t = x^2+bx+c,\:
{\textstyle\frac{dt}{dx} = 2x+b}\:
\right]$


(C.3).
$\displaystyle\int \frac{1}{x^2+bx+c} dx
= \begin{cases}
\frac{2}{\vert b^2-4c...
...b^2-4c\vert^\frac{1}{2}}\right)^{**}
& \mbox{ if $b^2 - 4c > 0$. }
\end{cases}$


$\left[
^{**}\mbox{ Use }\:
\frac{1}{x^2+bx+c} = \frac{4}{(2x+b)^2-(b^2-4c)}
\:\...
...:
{\textstyle\frac{dx}{dt} = \frac{\vert b^2-4c\vert^\frac{1}{2}}{2}}\:
\right]$


\begin{align*}
& \mbox{(a) }
\int \frac{1}{x^2 + x + 1} dx
= \frac{2}{\sqrt{3}}...
...2x+1}{\sqrt{3}}\right) + C
\hspace{0.8in}\fbox{(C.3)}\hspace{5.0in}
\end{align*}


\begin{align*}
& \mbox{(b) }
\int \frac{x}{x^4+x^2+1} dx
= \int \frac{x}{(x^2+\...
...ac{1}{\sqrt{3}} \tan^{-1}
\left(\frac{2x^2+1}{\sqrt{3}}\right) + C
\end{align*}


(C.4).
$\displaystyle\int \frac{1}{(1 + x^2)^{n+1}} dx
= \frac{x}{2n(1 + x^2)^n}
+ \frac{2n-1}{2n} \int \frac{1}{(1 + x^2)^n} dx^{***}$

$\displaystyle\int \frac{1}{(1 - x^2)^{n+1}} dx
= \frac{x}{2n(1 - x^2)^n}
+ \frac{2n-1}{2n} \int \frac{1}{(1 - x^2)^n} dx$


$\displaystyle\left[
^{***}\mbox{ Use }\:
\frac{d}{dx}\left(\frac{x}{(1+x^2)^n}\right)
= -\frac{2n-1}{(1+x^2)^n} + \frac{2n}{(1+x^2)^{n+1}}\:
\right]$


\begin{align*}
& \mbox{(a) }
\int \sec^5 x dx
= \int \frac{1}{\cos^5 x} dx
= \...
... x}
+ \frac{3\sin x}{8\cos^2 x} + \frac{3}{8}\tanh^{-1} \sin x + C
\end{align*}


\begin{align*}
& \mbox{(b) }
\int \frac{1}{(x^2 + 4)^3} dx
= \frac{1}{32} \int ...
...}
+ \frac{3x}{128(x^2 + 4)}
+ \frac{3}{256}\tan^{-1}\frac{x}{2} + C
\end{align*}



Department of Mathematics
Last modified: 2005-03-11