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Cheat Sheet

MATH 125

Table of contents
  1. Antiderivatives
  2. Techniques
  3. Trigonometric Identities
  4. Work
  5. Specific Integrals


\[\int 0 \: dx = C\] \[\int cf\left(x\right)\: dx=c\int f\left(x\right)dx\] \[\int x^n\: dx=\frac{x^{n+1}}{n+1}+C\left(n\ne -1\right)\] \[\int e^x\: dx=e^x+C\] \[\int \sin x\:dx=-\cos x+C\]

$\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$$

\[\int \cosh x\:dx=\sinh x+C\] \[\int \tan x \:dx = -\ln \| \cos(x) \| + C\] \[\int \sec x \:dx = \ln \| \tan(x) + \sec(x) \| + C\] \[\int \csc x \:dx = \ln \| \tan\left(\frac{x}{2}\right)\| + C\] \[\int \sin^2 x\:dx = \frac{1}{2} \left(x - \frac{1}{2} \sin(2x)\right) + C\] \[\int \cos^2 x\:dx = \frac{1}{2} \left(x + \frac{1}{2} \sin (2x) \right) + C\] \[\int \sin^2(x)\cos^2(x)\:dx = \frac{1}{8} \left(x - \frac{1}{4} \sin(4x) \right) + C\] \[\int b^xdx=\frac{b^x}{\ln b}+C\] \[\int \frac{1}{\sqrt{1-x^2}}\:dx=\sin ^{-1}x+C\] \[\int \frac{dx}{x^2 + a^2} = \frac{1}{a} \tan^{-1} \left\frac{x}{a}\right)\] \[\int \frac{dx}{\sqrt{a^2 - x^2}} = \sin^{-1} \left(\frac{x}{a}\right), a > 0\] \[\int \frac{dx}{x^2 - a^2} = \frac{1}{2a} \ln \left\| \frac{x-a}{x+a} \right\|\] \[\int \frac{dx}{\sqrt{x^2 \pm \a^2}} = \ln \left\| x + \sqrt{x^2 \pm a^2} \right\|\]


\(u\)-SubstitutionGiven \(\int f(x) dx\), 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 PartsWe 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 IdentitiesUse trigonometric identities to rewrite an integral into a more convenient form. See the below section on trigonometric identities.
Trigonometric SubstitutionLet \(x = a\sin\theta\), \(x = a\tan\theta\), 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 FractionsFactor 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

\[\sin^2 x + \cos^2 x = 1\] \[\sin^2 x = \frac{1}{2} \left(1 - \cos 2x\right)\] \[\cos^2 x = \frac{1}{2} \left(1 + \cos 2x\right)\] \[\sin x \cos x = \frac{1}{2} \sin 2x\] \[\sec^2 x = 1 + \tan^2 x\]


\[\text{Work} = \lim_{n\to\infty} \sum_{i=1}^n \text{Force} \times \text{Distance} = \int_a^b \text{Force} \times \text{Distance}\]

“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:
    • \[k = \text{density} = \text{weight per volume}\]
    • \[\text{Force} = k \cdot \text{volume} = k\cdot \text{horizontal slice area} \cdot \Delta y\]
    • \[\text{Distance} = \text{distance moved by slice} / a - y\]
    • \[\text{Work} = \int_0^b (a-y) k \cdot \text{horizontal slice area} \cdot dy\]

Specific Integrals

\[\int \sin ^2\left(x\right)dx=\frac{1}{2}\left(x-\frac{1}{2}\sin \left(2x\right)\right)+C\] \[\int \cos ^2\left(x\right)dx=\frac{1}{2}\left(x+\frac{1}{2}\sin \left(2x\right)\right)+C\] \[\int \tan \left(x\right)dx=-\ln \left|\cos \left(x\right)\right|+C\] \[\int \sec \left(x\right)dx=\ln \left|\tan \left(x\right)+\sec \left(x\right)\right|+C\] \[\int \csc \left(x\right)dx=\ln \left|\tan \left(\frac{x}{2}\right)\right|+C\] \[\int \cot \left(x\right)dx=\ln \left|\sin \left(x\right)\right|+C\]