Cheat Sheet
MATH 125
Table of contents
Antiderivatives
\[\int 0 \: dx = C\] \[\int cf\left(x\right)\: dx=c\int f\left(x\right)dx\] \[\int x^n\: dx=\frac{x^{n+1}}{n+1}+C\left(n\ne -1\right)\] \[\int e^x\: dx=e^x+C\] \[\int \sin x\:dx=-\cos x+C\]$\int \sec^2x:dx=\tan x+C$$
\[\int \sec x\:\tan x\:dx=\sec x+C\] \[\int \frac{1}{x^2+1}\:dx=\tan ^{-1}x+C\] \[\int \sinh x\:dx=\cosh x+C\] \[\int \frac{1}{x}\:dx=\ln \left|x\right|+C\] \[\int \cos x\:dx=\sin x+C\] \[\int \tan x\: dx = \ln \| \cos x \| + C\] \[\int \csc ^2x\:dx=-\cot x+C\]$\int \csc x\cot x:dx=-\csc x+C$$
\[\int \cosh x\:dx=\sinh x+C\] \[\int \tan x \:dx = -\ln \| \cos(x) \| + C\] \[\int \sec x \:dx = \ln \| \tan(x) + \sec(x) \| + C\] \[\int \csc x \:dx = \ln \| \tan\left(\frac{x}{2}\right)\| + C\] \[\int \sin^2 x\:dx = \frac{1}{2} \left(x - \frac{1}{2} \sin(2x)\right) + C\] \[\int \cos^2 x\:dx = \frac{1}{2} \left(x + \frac{1}{2} \sin (2x) \right) + C\] \[\int \sin^2(x)\cos^2(x)\:dx = \frac{1}{8} \left(x - \frac{1}{4} \sin(4x) \right) + C\] \[\int b^xdx=\frac{b^x}{\ln b}+C\] \[\int \frac{1}{\sqrt{1-x^2}}\:dx=\sin ^{-1}x+C\] \[\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left\frac{x}{a}\right)\] \[\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left(\frac{x}{a}\right), a > 0\] \[\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left\| \frac{x-a}{x+a} \right\|\] \[\int \frac{dx}{\sqrt{x^2 \pm \a^2}} = \ln \left\| x + \sqrt{x^2 \pm a^2} \right\|\]Techniques
| Technique | Description |
|---|---|
| \(u\)-Substitution | Given \(\int f(x) dx\), write \(u = g(x) \implies du = g'(x) dx\). We can rewrite the original integral as \(\int \frac{f(x)}{g'(x)} du \implies \int h(u) du\). The choice of \(u\) must allow for a more convenient rewriting such that \(h(u)\) can be easily integrated; after integration, plus in \(u\) for \(g(x)\). |
| Integration by Parts | We have that \(\frac{d}{dx} \left(f(x) \cdot g(x) \right) = f(x) \cdot g'(x) + f'(x) + g(x)\). This can be rewritten as \(\int f(x) g'(x) dx = f(x) \cdot g(x) - \int g(x) f'(x) dx\). When you are given an integral \(\int a(x) b(x) dx\), choose some \(u = a(x)\) and \(dv = b(x) dx\); this leads to \(du = a'(x) dx\) and \(v = \int b(x)\). We have that \(\int u dv = uv - \int v du\). |
| Trigonometric Identities | Use trigonometric identities to rewrite an integral into a more convenient form. See the below section on trigonometric identities. |
| Trigonometric Substitution | Let \(x = a\sin\theta\), \(x = a\tan\theta\), or \(x=a\sec\theta\) for integrals containing \(\sqrt{a^2 - x^2}\), \(\sqrt{a^2 + x^2}\), or \(\sqrt{x^2 - a^2\)} somewhere in the integrand. Substitute and simplify to get rid of the square root. To get the solution in terms of \(x\) again from \(\theta\), use the \triangle trick’. |
| Partial Fractions | Factor the denominator into a sum of terms of linear or quadratic denominators. Integrate each of these individually, either by using the \(\tan^{-1}\) identity or \(u\)-substitutions. |
Trigonometric Identities
\[\sin^2 x + \cos^2 x = 1\] \[\sin^2 x = \frac{1}{2} \left(1 - \cos 2x\right)\] \[\cos^2 x = \frac{1}{2} \left(1 + \cos 2x\right)\] \[\sin x \cos x = \frac{1}{2} \sin 2x\] \[\sec^2 x = 1 + \tan^2 x\]Work
\[\text{Work} = \lim_{n\to\infty} \sum_{i=1}^n \text{Force} \times \text{Distance} = \int_a^b \text{Force} \times \text{Distance}\]“Leaky Bucket” Problem: the pattern for force is given or we can find it. The force changes every small moment \(\text{distance} = \Delta x\) as the object is moved. We have:
- \[\text{Force} = f(x_i)\]
- \[\text{Distance} = \Delta x\]
- \[\text{Work} = \int_a^b f(x) dx\]
- Leaking at a constant rate \(f(X) = mx+b\) or \(f(x) = \text{force}\)
“Stack of Books”/Chain/Pumping Problem: we find the weight of a slice at a given height and find the formula for the distance that slice would move, then integrate across the length of the object.
- For chain problems:
- \[k = \text{density} = \text{force per distance}\]
- \[\text{Force} = \text{Weight of slice} = k\Delta x\]
- \[\text{Distance} = \text{distance moved by slice} / x\]
- \[\text{Work} = \int_0^b x k dx\]
- For pumping problems:
- \[k = \text{density} = \text{weight per volume}\]
- \[\text{Force} = k \cdot \text{volume} = k\cdot \text{horizontal slice area} \cdot \Delta y\]
- \[\text{Distance} = \text{distance moved by slice} / a - y\]
- \[\text{Work} = \int_0^b (a-y) k \cdot \text{horizontal slice area} \cdot dy\]