Summary of Integration Techniques

Standard forms

For those with two columns, you can see the right column as an extension of the left column. (The $\frac{1}{p}$ comes from what I call “reverse of Chain Rule”.) After forming a link this way, you only have to remember the left column.

$\int_{}^{}{adx} = ax + C$
$\int_{}^{}{x^{n}dx} = \frac{x^{n + 1}}{n + 1} + C,\text{ }(n \neq - 1)$	$$\int_{}^{}{(px + q)^{n}dx} = \frac{1}{p}\frac{(px + q)^{n + 1}}{n + 1} + C,\text{ }(n \neq - 1)$$
$\int_{}^{}{e^{x}dx} = e^{x} + C$	$$\int_{}^{}{e^{(px + q)}dx} = \frac{1}{p}e^{px + q} + C$$
$\int_{}^{}{k^{x}dx} = \frac{k^{x}}{\ln k} + C$	$$\int_{}^{}{k^{(px + q)}dx} = \frac{k^{px + q}}{p\ln k} + C$$
$\int_{}^{}{\frac{1}{x}dx} = \ln|x| + C$	$$\int_{}^{}{\frac{1}{px + q}dx} = \frac{1}{p}\ln|px + q| + C$$
$\int_{}^{}{\cos xdx = \sin x + C}$	$$\int_{}^{}{\cos(px + q)dx} = \frac{1}{p}\sin(px + q) + C$$
$\int_{}^{}{\sin xdx} = - \cos x + C$	$$\int_{}^{}{\sin(px + q)dx} = - \frac{1}{p}\cos(px + q) + C$$
$\int_{}^{}{\sec^{2}xdx} = \tan x + C$	$$\int_{}^{}{\sec^{2}(px + q)dx = \frac{1}{p}\tan(px + q) + C}$$
$\int_{}^{}{\text{cose}\text{c}^{2}xdx = - \cot x + C}$	$$\int_{}^{}{\text{cose}\text{c}^{2}(px + q)dx = - \frac{1}{p}\cot(px + q) + C}$$
$\int_{}^{}{\sec x\tan x dx} = \sec x + C$	$$\int_{}^{}{\sec(px + q)\tan(px + q)dx} = \frac{1}{p}\sec(px + q) + C$$
$\int_{}^{}{{cosec}x\cot xdx} = - {cosec}x + C$	$$\int_{}^{}{{cosec}(px + q)\cot(px + q)dx} = - \frac{1}{p}{cosec}(px + q) + C$$
$\int_{}^{}{\frac{1}{\sqrt{1 - x^{2}}}dx} = \sin^{- 1}x + C$	$$\int_{}^{}{\frac{1}{\sqrt{a^{2} - (px + q)^{2}}}dx} = {\frac{1}{p}\sin}^{- 1}\frac{px + q}{a} + C$$
$\int_{}^{}{\frac{1}{1 + x^{2}}dx} = \tan^{- 1}x + C$	$$\int_{}^{}{\frac{1}{a^{2} + (px + q)^{2}}dx} = {\frac{1}{pa}\tan}^{- 1}\frac{px + q}{a} + C$$

Expressions containing $\mathbf{f(x)}$ and $\text{f}\text{ }^{\mathbf{'}}\mathbf{(x)}$

$\int_{}^{}{\text{f}\text{ }'(x)\left\lbrack f(x) \right\rbrack^{n}dx} = \frac{\left\lbrack f(x) \right\rbrack^{n + 1}}{n + 1} + C,$ n ≠ −1
$\int_{}^{}{\frac{f'(x)}{f(x)}dx} = \ln\left| f(x) \right| + C$	Use this for: $\int_{}^{}{\tan x dx}$ and $\int_{}^{}{\cot xdx}$
$\int_{}^{}{\text{f}\text{ }'(x)e^{f(x)}dx} = e^{f(x)} + C$
$\int_{}^{}{\text{f}\text{ }'(x)\sin f(x)dx} = - \cos\ f(x) + C$

Techniques of integration

Other trigonometric functions

Integrals	How?
$\int_{}^{}{\sin^{n}(kx)\text{ d}x}\text{ or }\int_{}^{}{\cos^{n}(kx)\text{ d}x}$, where n is even	$${\sin^{2}x = \frac{1}{2}\left\lbrack 1 - \cos(2x) \right\rbrack }{\cos^{2}x = \frac{1}{2}\left\lbrack 1 + \cos(2x) \right\rbrack}$$
$\int_{}^{}{\sin^{n}(kx)\ }dx$ $$\int_{}^{}{\cos^{n}(kx)\ }dx$$ $\int_{}^{}{\sin^{m}(kx)\cos^{n}(kx)\ }dx$ $\int_{}^{}{\cos^{m}(kx)\sin^{n}(kx)\ }dx$ where n is odd and m is integer	Pull out one copy of the one with odd power E.g. $$\int_{}^{}{\cos^{2}x\sin^{3}x\ }dx = \int_{}^{}{\cos^{2}x\left( 1 - \cos^{2}x \right)\sin x\ }dx$$ $$= \int_{}^{}{(\cos^{2}x - \cos^{4}x)\sin x\ }dx$$ Note: If both m and n are even, use $\cos^{2}x = 1 - \sin^{2}x$ or $\sin^{2}x = 1 - \cos^{2}x$ E.g. $$\int_{}^{}{\cos^{2}x\sin^{2}x\ }dx = \int_{}^{}{\cos^{2}x\left( 1 - \cos^{2}x \right)\ }dx$$ $$= \int_{}^{}{\cos^{2}x - \cos^{4}x\ }dx$$
$$\int_{}^{}{\tan^{n}(kx)\text{ d}x}\text{ or }\int_{}^{}{\cot^{n}(kx)\text{ d}x}$$	n = 1: Use tan (kx) = $\frac{\sin(kx)}{\cos(kx)}$ or cot (kx) = $\frac{\cos(kx)}{\sin(kx)}$, and then $\int_{}^{}{\frac{f'(x)}{f(x)}dx} = \ln\left| f(x) \right| + C$ n ≥ 2: Pull out one copy of: $\tan^{2}x = \sec^{2}x - 1$$$\cot^{2}x = \text{cose}\text{c}^{2}x - 1$$
$$\int_{}^{}{\sin(kx)\cos(mx)\text{ d}x}$$ $$\int_{}^{}{\cos(kx)\cos(mx)\text{ d}x}$$ $$\int_{}^{}{\sin(kx)\sin(mx)\text{ d}x}$$	$$\sin A\cos B = \frac{1}{2}\lbrack\sin(A + B) + \sin(A - B)\rbrack$$ $${\cos A\sin B = \frac{1}{2}\lbrack\sin(A + B) - \sin(A - B)\rbrack}$$ $${\cos A\cos B = \frac{1}{2}\lbrack\cos(A - B) + \cos(A + B)\rbrack}$$ $${\sin A\sin B = \frac{1}{2}\lbrack\cos(A - B) - \cos(A + B)\rbrack}$$

Rational algebraic functions and similar ones

Steps
Improper? If yes, use long division or juggling.
Can each part be integrated by standard forms?
If not, would splitting the numerator help?
If not, would partial fractions do the trick?
Integrals	How?
$\int_{}^{}{\frac{1}{\sqrt{ax^{2} + bx + c}}dx}$	Complete the square for $ax^{2} + bx + c$
$\int_{}^{}{\frac{1}{ax^{2} + bx + c}dx}$	If $ax^{2} + bx + c$ can be factorised, factorise and use partial-fractions method. If not, complete the square.
$\int_{}^{}{\frac{px + q}{\sqrt{ax^{2} + bx + c}}dx}$	$$px + q \equiv A\frac{d}{dx}(ax^{2} + bx + c) + B$$ $\equiv A(2ax + b) + B$ where A and B are constants to be determined
$\int_{}^{}{\frac{1}{ax^{2} + bx + c}dx}$

By substitution

$$\int_{}^{}{g\left( f(x) \right)dx} = \int_{}^{}{g(u)\frac{dx}{du}du},\ where\ u = f(x)$$

(Note: The required substitution would always be given in the question.)

By parts

$$\int_{}^{}{udv\text{ } = \text{ }uv - \int_{}^{}{v\text{ d}u}}$$

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