Summary Table Of Integrals

Indefinite Integral $$\begin{align*} \int u^\alpha\; du &={u^{\alpha+1}\over\alpha+1}+c, \quad \alpha\ne-1 \\ &\\ \int{du\over u} &=\ln|u|+c \\ &\\  \int\cos u\; du&=\sin u+c \\ &\\ \int \sin u\; du&=-\cos u+c \\ &\\ \int \tan u\; du&=-\ln|\cos u|+c \\ &\\ \int \cot u\; du&=\ln|\sin u|+c \\ &\\ \int \sec^2 u\; du&=\tan u+c \\ &\\ \int \csc^2 u\; du&=-\cot u+c \\ &\\ \int \sec u\; du&=\ln|\sec u+\tan u|+c \\ &\\ \int \csc( {u})du  &= \ln|\csc( {u}) - \cot( {u})| + c\\ &\\ \int\cos^2 u\; du&={u\over2}+{1\over4}\sin2u+c \\ &\\ \int\sin^2 u\; du&={u\over2}-{1\over4}\sin2u+c \\ &\\ \int {du\over 1+u^2}\; du&=\tan^{-1}u+c \\ &\\ \int {du\over\sqrt{1-u^2}}\; du&=\sin^{-1}u+c \\ &\\ \int {1\over u^2-1}\; du&={1\over2}\ln\left|u-1\over u+1\right|+c \\ &\\ \end{align*}$$	Integration Rules $\displaystyle \int (A{f(x)} + B{g(x)} dx = A \int {f(x)}dx + B \int {g(x)} dx$  $\displaystyle \int {f'(} {g(x)} {)} {g'(x)}dx = {f(} {g(x)} {)} + C$ Definite Integral $\displaystyle \int^a_af(x)\,dx=0$ $\displaystyle \int^a_bf(x)\,dx=-\displaystyle \int^b_af(x)dx$ $\displaystyle \int^b_a[f(x)+g(x)]\,dx=\displaystyle \int^b_af(x)dx+\displaystyle \int^b_ag(x)dx$ $\displaystyle \int^b_a[f(x)-g(x)]\,dx=\displaystyle \int^b_af(x)dx-\displaystyle \int^b_ag(x)dx$ $\displaystyle \int^b_acf(x)\,dx=c\displaystyle \int^b_af(x)$ for constant c. $\displaystyle \int^b_af(x)\,dx=\displaystyle \int^c_af(x)\,dx+\displaystyle \int^b_cf(x)\,dx$ Although this formula normally applies when c is between a and b, the formula holds for all values of $a$, $b$, and $c$, provided $f(x)$ is integrable on the largest interval. Let $f$ be continuous on $[a,b]$ and let $F$ be any anti-derivative of $f$. Then $\int_a^b f(x)dx = F(b) - F(a).$  Reduction formulas $$\int \sin^n (x) \ dx =- \frac{1}{n}\sin^{n-1}x \cos x + \frac{n-1}{n}\int \sin^{n-2}x\, dx .$$ $$\int \cos^n (x) \ dx = \frac{1}{n}\cos^{n-1}x \sin x + \frac{n-1}{n}\int \cos^{n-2}x\, dx .$$ $$\int \tan^n (x) \ dx = \frac{1}{n-1}\tan^{n-1}x - \int \tan^{n-2}x\ dx , n\ne 1.$$ $$\int \sec^n(x) \ dx = \frac{1}{n-1}\sec^{n-2}x\tan x + \frac{n-2}{n-1}\int \sec^{n-2}x\ dx , n \geq 2.$$
Integration by parts $$\begin{align*} \int u\; dv&=uv-\int v\; du \\ &\\ \int u\cos u\; du&=u\sin u +\cos u+c \\ &\\ \int u\sin u\; du&=-u\cos u +\sin u+c \\ &\\ \int ue^u\; du&=ue^{u}-e^{u} +c \\ &\\ \int e^{\lambda u}\cos\omega u\; du&={e^{\lambda u}(\lambda\cos\omega u+\omega\sin\omega u)\over \lambda^2+\omega^2}+c \\ &\\ \int e^{\lambda u}\sin\omega u\; du&={e^{\lambda u}(\lambda\sin\omega u-\omega\cos\omega u)\over\lambda^2+\omega^2}+c \\ &\\ \int \ln|u|\; du&=u\ln|u|-u+c \\ &\\ \int u\ln|u|\; du&={u^2\ln|u|\over2}-{u^2\over4}+c \\ &\\ \int\cos\omega_1u\cos\omega_2u\,du&={\sin(\omega_1+\omega_2)u\over2(\omega_1+\omega_2)} \\ &\\ &+{\sin(\omega_1-\omega_2)u\over2(\omega_1-\omega_2)}+c\quad (\omega_1\ne\pm\omega_2) \\ &\\ \int\sin\omega_1 u\sin\omega_2 u\,du&=-{\sin(\omega_1+\omega_2)u\over2(\omega_1+\omega_2)}+{\sin(\omega_1-\omega_2)u\over2(\omega_1-\omega_2)}+c \quad (\omega_1\ne\pm\omega_2) \\ &\\ \int\sin\omega_1u\cos\omega_2 u\,du&=-{\cos(\omega_1+\omega_2)u\over2(\omega_1+\omega_2)}-{\cos(\omega_1-\omega_2)u\over2(\omega_1-\omega_2)}+c\quad (\omega_1\ne\pm\omega_2)\end{align*}$$	Improper Integrals Let $f(x)$ be continuous over an interval of the form $[a,+\infty)$. Then $$\int ^{+\infty}_af(x)dx=\lim_{t \to +\infty}\int ^t_af(x)dx, \label{improper1}$$ provided this limit exists.  Let $f(x)$ be continuous over an interval of the form $(-\infty,b]$. Then $$\int ^b_{-\infty}f(x)dx=\lim_{t \to -\infty}\int ^b_tf(x)dx, \label{improper2}$$ provided this limit exists. In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge. Let $f(x)$ be continuous over $(-\infty,+\infty)$. Then $$\int ^{+\infty}_{-\infty}f(x)dx=\int ^0_{-\infty}f(x)dx+\int ^{+\infty}_0f(x)dx, \label{improper3}$$ provided that $\int ^0_{-\infty}f(x)dx$ and $\int ^{+\infty}_0f(x)dx$ both converge. If either of these two integrals diverge, then $\int ^{+\infty}_{-\infty}f(x)dx$ diverges. (It can be shown that, in fact, $\int ^{+\infty}_{-\infty}f(x)dx=\int ^a_{-\infty}f(x)dx+\int ^{+\infty}_af(x)dx$ for any value of a.). Let $f(x)$ be continuous over $[a,b)$. Then, $$\int ^b_af(x)dx=\lim_{t \to b^-}\int ^t_af(x)dx.$$ Let $f(x)$ be continuous over $(a,b]$. Then, $$\int ^b_af(x)dx=\lim_{t \to a^+}\int ^b_tf(x)dx.$$ In each case, if the limit exists, then the improper integral is said to converge. If the limit does not exist, then the improper integral is said to diverge.  If $f(x)$ is continuous over $[a,b]$ except at a point $c$ in $(a,b)$, then $$\int ^b_af(x)dx=\int ^c_af(x)dx+\int ^b_cf(x)dx,$$ provided both $\int ^c_af(x)dx$ and $\int ^b_cf(x)dx$ converge. If either of these integrals diverges, then $\int ^b_af(x)dx$ diverges.