Provided formulas
List of provided integrals (arbitrary constants omitted)
$\displaystyle \int \frac{1}{x^2+a^2} \; \mathrm{d}x = \frac{1}{a} \tan^{-1} \left ( \frac{x}{a} \right ) \\
\displaystyle \int \frac{1}{\sqrt{a^2 -x^2}} \; \mathrm{d}x = \sin^{-1} \left ( \frac{x}{a} \right ) \qquad \mathrm{for} \; |x| < a \\
\displaystyle \int \frac{1}{x^2-a^2} \; \mathrm{d}x = \frac{1}{2a} \ln \left ( \frac{x-a}{x+a} \right ) \qquad \mathrm{for} \; x > a \\
\displaystyle \int \frac{1}{a^2-x^2} \; \mathrm{d}x = \frac{1}{2a} \ln \left ( \frac{a+x}{a-x} \right ) \qquad \mathrm{for} \; |x| < a$
Remark: For most cases (i.e. in the event that the domain is uncertain), we use
$\displaystyle \int \frac{1}{x^2-a^2} \; \mathrm{d}x = \frac{1}{2a} \ln \left | \frac{x-a}{x+1} \right |$.
$\displaystyle \int \tan x \; \mathrm{d}x = \ln ( \mathrm{sec} \; x) \qquad \textrm{for } |x| < \frac{\pi}{2}\\
\displaystyle \int \mathrm{cot} \; x \; \mathrm{d}x = \ln ( \sin x) \qquad \textrm{for } 0 < x < \pi \\
\displaystyle \int \mathrm{cosec} \; x \; \mathrm{d}x = -\ln ( \mathrm{cosec} \; x + \mathrm{cot} \; x ) \qquad \textrm{for } 0 < x < \pi \\
\displaystyle \int \mathrm{sec} \; x \; \mathrm{d}x = \ln ( \mathrm{sec} \; x + \tan x ) \qquad \textrm{for } |x| < \frac{\pi}{2} $
Selected list of trigonometric identities
$\cos 2A \equiv \cos^2 A - \sin^2 A \equiv 2 \cos^2 A - 1 \equiv 1 - 2\sin^2 A \\
\sin P + \sin Q \equiv 2 \sin \frac{1}{2} (P+Q) \cos \frac{1}{2} (P-Q) \\
\sin P - \sin Q \equiv 2 \cos \frac{1}{2} (P+Q) \sin \frac{1}{2} (P-Q) \\
\cos P + \cos Q \equiv 2 \cos \frac{1}{2} (P+Q) \cos \frac{1}{2} (P-Q) \\
\cos P - \cos Q \equiv -2 \sin \frac{1}{2} (P+Q) \sin \frac{1}{2} (P-Q) $
Formulas that are not provided
$f'(x)$ type integrals (arbitrary constants omitted)
$\displaystyle \int \frac{f'(x)}{f(x)} \; \mathrm{d}x = \ln | f(x) | \\
\displaystyle \int f'(x) \big ( f(x) \big )^n \; \mathrm{d}x = \frac{ \big (f(x) \big )^{n+1}}{n+1} \qquad \mathrm{for} \; n \neq -1 \\
\displaystyle \int f'(x) \; \mathrm{e}^{f(x)} \; \mathrm{d}x = \mathrm{e}^{f(x)}$
Integration by parts
$\displaystyle \int u \frac{\mathrm{d}v}{\mathrm{d}x} \; \mathrm{d}x = uv - \int v \frac{\mathrm{d}u}{\mathrm{d}x} \; \mathrm{d}x$
Integration by Substitution
$\displaystyle \int f(x) \; \mathrm{d}x = \int f \big ( g(u) \big ) \frac{\mathrm{d}x}{\mathrm{d}u} \; \mathrm{d}x \qquad \mathrm{where} \; x = g(u)$
The following questions illustrate the key concepts needed in this topic. See if you can answer them.
Click on the title for detailed discussion of how to tackle them.
Click on or hover over the question to show the answer.
$\displaystyle \int \frac{1}{2x-3} \; \mathrm{d}x$
$\displaystyle \frac{\ln | 2x-3 |}{2} + C$
$\displaystyle \int \cos (3-2x) \; \mathrm{d}x $
$\displaystyle \frac{\sin(3-2x)}{-2} + C $
$\displaystyle \int \frac{x}{x^2 - 5} \; \mathrm{d}x$
$\displaystyle \frac{\ln \left | x^2 - 5 \right | }{2} + C $
$\displaystyle \int \frac{x}{\sqrt{x^2 - 5}} \; \mathrm{d}x$
$\displaystyle \sqrt{x^2-5} + C $
$\displaystyle \int x \; \mathrm{e}^{x^2 - 5} \; \mathrm{d}x$
$\displaystyle \frac{\mathrm{e}^{x^2 - 5}}{2} + C $
$\displaystyle \int \frac{1}{x^2 + 4} \; \mathrm{d}x $
$\displaystyle \frac{1}{2} \tan^{-1} \left ( \frac{x}{2} \right) + C$
$\displaystyle \int \frac{1}{\sqrt{5 -x^2}} \; \mathrm{d}x$
$\displaystyle \sin^{-1} \left ( \frac{x}{\sqrt{5}} \right) + C$
$\displaystyle \int \frac{1}{9 - 4 x^2} \; \mathrm{d}x $
$\displaystyle \frac{1}{12} \ln \left | \frac{3+2x}{3-2x} \right | + C$
$\displaystyle \int \frac{1}{x^2 + 2x - 4} \; \mathrm{d}x $
$\displaystyle \frac{1}{2\sqrt{5}} \ln \left | \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right | + C$
$\displaystyle \int \frac{2x}{x^2 + 2x - 4} \; \mathrm{d}x $
$\displaystyle \ln |x^2 +2x - 4| - \frac{1}{\sqrt{5}} \ln \left | \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right | + C$
$\displaystyle \int \frac{x^2}{x^2 + 2x - 4}\; \mathrm{d}x $
$\displaystyle x - \ln |x^2+2x-4| + \frac{3}{\sqrt{5}} \ln \left | \frac{x+1-\sqrt{5}}{x+1+\sqrt{5}} \right | + C$
$\displaystyle \int \frac{1}{x^3 - x} \; \mathrm{d}x $
$\displaystyle \frac{1}{2} \ln |x-1| - \ln |x| + \frac{1}{2} \ln |x+1| + C$
$\displaystyle \int \tan^2 x \; \mathrm{d}x$
$\tan x - x + C$
$\displaystyle \int \sin^2 x \; \mathrm{d}x$
$\displaystyle \frac{x}{2}-\frac{\sin 2x}{4} + C$
$\displaystyle \int \sin 5x \cos 3x \; \mathrm{d} x$
$\displaystyle -\frac{\cos 8x}{16} - \frac{\cos 2x}{4} + C $
$\displaystyle \int x \, \mathrm{e}^{3x} \; \mathrm{d}x$
$\displaystyle \frac{x \, \mathrm{e}^{3x}}{3} - \frac{\mathrm{e}^{3x}}{9} + C$
$\displaystyle \int \ln x \; \mathrm{d}x$
$x \ln x - x + C$
$\displaystyle \int \mathrm{e}^{x} \sin 2x \; \mathrm{d}x$
$\displaystyle \frac{\mathrm{e}^x \sin 2x - 2 \mathrm{e}^x \cos 2x}{5} + C$
$\displaystyle \int x \sqrt{2x-3} \; \mathrm{d}x$
using the substitution $u=2x-3$ $\displaystyle \frac{1}{10} (2x-3)^{\frac{5}{2}} + \frac{1}{2} (2x-3)^{\frac{3}{2}} + C$
using the substitution $u=2x-3$ $\displaystyle \frac{1}{10} (2x-3)^{\frac{5}{2}} + \frac{1}{2} (2x-3)^{\frac{3}{2}} + C$
$\displaystyle \int \sqrt{1-x^2} \; \mathrm{d}x$
using the substitution $x = \sin \theta$ $\displaystyle \frac{x\sqrt{1-x^2}+\sin^{-1}x}{2} + C$
using the substitution $x = \sin \theta$ $\displaystyle \frac{x\sqrt{1-x^2}+\sin^{-1}x}{2} + C$