Integration by Parts
\(\int u\,dv = uv - \int v\,du\)
LIATE rule for choosing \(u\): Logarithm, Inverse trig, Algebraic, Trig, Exponential.
Trigonometric Substitution
- \(\sqrt{a^2-x^2}\): let \(x=a\sin\theta\)
- \(\sqrt{a^2+x^2}\): let \(x=a\tan\theta\)
- \(\sqrt{x^2-a^2}\): let \(x=a\sec\theta\)
Partial Fractions
Decompose rational functions. For \(\frac{P(x)}{Q(x)}\) with \(\deg P < \deg Q\), split by linear and irreducible quadratic factors.
Improper Integrals
\(\int_1^\infty \frac{1}{x^p}\,dx\) converges iff \(p>1\).
Use limits: \(\lim_{b\to\infty}\int_a^b f(x)\,dx\)
Sequences & Series
Geometric series: \(\sum_{n=0}^\infty ar^n = \frac{a}{1-r}\), \(|r|<1\).
Harmonic series \(\sum \frac{1}{n}\) diverges.
p-series \(\sum \frac{1}{n^p}\) converges iff \(p>1\).
Convergence Tests
- Ratio test: \(L = \lim\left|\frac{a_{n+1}}{a_n}\right|\)
- Root test: \(L = \lim\sqrt[n]{|a_n|}\)
- Comparison: compare with known series
- Alternating: decrease to zero → converges
Power Series & Radius
\(\sum c_n(x-a)^n\), radius \(R = \frac{1}{\limsup\sqrt[n]{|c_n|}}\)
Always check endpoints separately.
Taylor & Maclaurin Series
- \(e^x = \sum_{n=0}^\infty \frac{x^n}{n!}\)
- \(\sin x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!}\)
- \(\cos x = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!}\)
- \(\ln(1+x) = \sum_{n=1}^\infty \frac{(-1)^{n+1}x^n}{n}\), \(|x|\le1, x\ne-1\)
- Integral of \(\sec\theta\): \(\int\sec\theta\,d\theta = \ln|\sec\theta+\tan\theta|+C\)
- Integral of \(\frac{1}{x^2+a^2}\): \(\frac{1}{a}\arctan\!\frac{x}{a}+C\)
- Alternating Series Estimation: \(|S-S_n|\le b_{n+1}\)
- Taylor remainder: \(|R_n(x)|\le \frac{M|x-a|^{n+1}}{(n+1)!}\)
- Arc length: \(L=\int_a^b\sqrt{1+[f'(x)]^2}\,dx\)
- Surface area: \(S=2\pi\int_a^b f(x)\sqrt{1+[f'(x)]^2}\,dx\)
Evaluate \(\int x e^x\,dx\).
Let \(u=x\), \(dv=e^x dx\) → \(du=dx\), \(v=e^x\).
\(\int x e^x\,dx = xe^x - \int e^x\,dx = xe^x - e^x + C = e^x(x-1)+C\) ✓
Test \(\sum_{n=1}^\infty \frac{n!}{n^n}\) for convergence.
\(L=\lim_{n\to\infty}\frac{(n+1)!/(n+1)^{n+1}}{n!/n^n}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim\frac{1}{(1+1/n)^n}=\frac{1}{e}<1\)
By Ratio Test: converges. ✓