# Cheat Sheet

MATH 125

## 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

TechniqueDescription
$$u$$-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

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

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

## Specific Integrals

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