Important Results

Two substitution patterns come up so often that it pays to know them as ready-made formulas rather than re-deriving them every time.

Result (1). $\displaystyle\int\dfrac{f'(x)}{f(x)}\,dx=\log|f(x)|+c$ — whenever the numerator is (a constant multiple of) the exact derivative of the denominator, the integral is the log of the denominator's absolute value.

Proof (own words). Put $u=f(x)$, so $du=f'(x)\,dx$; the integral becomes $\displaystyle\int\dfrac{du}{u}=\log|u|+c$, and re-substituting $u=f(x)$ gives the stated result.

Result (2). $\displaystyle\int f'(x)\,[f(x)]^n\,dx=\dfrac{[f(x)]^{n+1}}{n+1}+c,\quad n\ne-1$ — a "chunk raised to a power" times the exact derivative of that chunk integrates by the ordinary power rule applied to the chunk.

Proof (own words). Again put $u=f(x)$, $du=f'(x)\,dx$; the integral becomes $\displaystyle\int u^n\,du=\dfrac{u^{n+1}}{n+1}+c$, and re-substituting gives the result. (Result (1) is really the $n=-1$ case of this same pattern, which is exactly why that case has to be stated separately — the power rule itself is undefined at $n=-1$.)

The four standard log-form results, each a direct application of Result (1):

Writing $\tan x=\dfrac{\sin x}{\cos x}$ and setting $u=\cos x$ (so $du=-\sin x\,dx$) gives $\displaystyle\int\tan x\,dx=-\log|\cos x|+c=\log|\sec x|+c$. Writing $\cot x=\dfrac{\cos x}{\sin x}$ and setting $u=\sin x$ gives $\displaystyle\int\cot x\,dx=\log|\sin x|+c$. For $\csc x$, multiply and divide by $(\csc x-\cot x)$: $\csc x=\dfrac{\csc x(\csc x-\cot x)}{\csc x-\cot x}=\dfrac{\csc^2x-\csc x\cot x}{\csc x-\cot x}$, whose numerator is exactly the derivative of the denominator, so by Result (1), $\displaystyle\int\csc x\,dx=\log|\csc x-\cot x|+c$. Symmetrically, multiplying and dividing $\sec x$ by $(\sec x+\tan x)$ gives numerator $\sec^2x+\sec x\tan x$, the derivative of $\sec x+\tan x$, so $\displaystyle\int\sec x\,dx=\log|\sec x+\tan x|+c$.

$\int\csc x\,dx=\log|\csc x-\cot x|+c,\qquad \int\sec x\,dx=\log|\sec x+\tan x|+c.$ …