Must memorize

Only the essentials: the things you can't work out quickly in an exam. In this course log⁡\log means the natural log, ln⁡\ln. Practice them as flashcards.

Integration formulas

The antiderivatives and rules every technique question leans on.
  • ddx∫g(x)h(x)f(t) dt=f(h(x)) h′(x)−f(g(x)) g′(x)\dfrac{d}{dx}\int_{g(x)}^{h(x)} f(t)\,dt = f(h(x))\,h'(x) - f(g(x))\,g'(x)
  • ∫abf(g(x)) g′(x) dx=∫g(a)g(b)f(u) du\int_a^b f(g(x))\,g'(x)\,dx = \int_{g(a)}^{g(b)} f(u)\,du
    u=g(x)u = g(x): change the limits too.
  • ∫u dv=uv−∫v du\int u\,dv = uv - \int v\,du
    Pick uu so it gets simpler when you differentiate it.
  • ∫sec⁡x dx=log⁡∣sec⁡x+tan⁡x∣+C\int\sec x\,dx = \log|\sec x + \tan x| + C
  • ∫tan⁡x dx=log⁡∣sec⁡x∣+C\int\tan x\,dx = \log|\sec x| + C
  • ddxsec⁡x=sec⁡xtan⁡x\dfrac{d}{dx}\sec x = \sec x\tan x
    So with u=sec⁡xu = \sec x, du=sec⁡xtan⁡x dxdu = \sec x\tan x\,dx.
  • ∫dxx2+a2=1aarctan⁡xa+C\int\dfrac{dx}{x^2 + a^2} = \dfrac{1}{a}\arctan\dfrac{x}{a} + C
    Gives the arctan part of an in partial fractions.
  • ∫dxa2−x2=arcsin⁡xa+C\int \dfrac{dx}{\sqrt{a^2 - x^2}} = \arcsin\dfrac{x}{a} + C
    For a>0a > 0.

Trig identities and substitutions

The identities that rewrite leftover powers, and the substitutions that remove roots.
  • sin⁡2x=12(1−cos⁡2x)\sin^2 x = \tfrac12(1 - \cos 2x)
  • cos⁡2x=12(1+cos⁡2x)\cos^2 x = \tfrac12(1 + \cos 2x)
  • tan⁡2x=sec⁡2x−1\tan^2 x = \sec^2 x - 1
    Divide sin⁡2x+cos⁡2x=1\sin^2 x + \cos^2 x = 1 by cos⁡2x\cos^2 x.
  • For a2−x2\text{For } \sqrt{a^2 - x^2}
    x=asin⁡θx = a\sin\theta
    The root becomes acos⁡θa\cos\theta.
  • For a2+x2\text{For } \sqrt{a^2 + x^2}
    x=atan⁡θx = a\tan\theta
    The root becomes asec⁡θa\sec\theta.
  • For x2−a2\text{For } \sqrt{x^2 - a^2}
    x=asec⁡θx = a\sec\theta
    The root becomes atan⁡θa\tan\theta.

Riemann sums and numerical integration

The two sums a Riemann-sum limit needs, then the three rules and their error bounds. Here Δx=b−an\Delta x = \dfrac{b-a}{n}.
  • ∑i=1ni=n(n+1)2\sum_{i=1}^{n} i = \dfrac{n(n+1)}{2}
  • ∑i=1ni2=n(n+1)(2n+1)6\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}
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