Each standard differentiation result, read in reverse, gives a standard integral. For the linear form \(ax+b\), differentiating multiplies by \(a\) — so integrating divides by \(a\).
Trigonometric integrals (angles in radians)
| Integral | Result | Example |
|---|---|---|
| \(\displaystyle\int\sin x\,\mathrm{d}x\) | \(-\cos x+c\) | \(\displaystyle\int 3\sin x\,\mathrm{d}x=-3\cos x+c\) |
| \(\displaystyle\int\cos x\,\mathrm{d}x\) | \(\sin x+c\) | \(\displaystyle\int\dfrac{\cos x}{2}\,\mathrm{d}x=\dfrac12\sin x+c\) |
| \(\displaystyle\int\sec^2 x\,\mathrm{d}x\) | \(\tan x+c\) | \(\displaystyle\int 4\sec^2 x\,\mathrm{d}x=4\tan x+c\) |
| \(\displaystyle\int\sin(ax+b)\,\mathrm{d}x\) | \(-\dfrac{\cos(ax+b)}{a}+c\) | \(\displaystyle\int\sin(3x+2)\,\mathrm{d}x=-\dfrac{\cos(3x+2)}{3}+c\) |
| \(\displaystyle\int\cos(ax+b)\,\mathrm{d}x\) | \(\dfrac{\sin(ax+b)}{a}+c\) | \(\displaystyle\int\cos(5x-1)\,\mathrm{d}x=\dfrac{\sin(5x-1)}{5}+c\) |
| \(\displaystyle\int\sec^2(ax+b)\,\mathrm{d}x\) | \(\dfrac{\tan(ax+b)}{a}+c\) | \(\displaystyle\int 2\sec^2(4x)\,\mathrm{d}x=\dfrac{\tan(4x)}{2}+c\) |
Watch the sign: \(\sin\) integrates to \(-\cos\), while \(\cos\) integrates to \(+\sin\).
Exponential integrals
| Integral | Result | Example |
|---|---|---|
| \(\displaystyle\int e^x\,\mathrm{d}x\) | \(e^x+c\) | \(\displaystyle\int 7e^x\,\mathrm{d}x=7e^x+c\) |
| \(\displaystyle\int e^{ax+b}\,\mathrm{d}x\) | \(\dfrac{e^{ax+b}}{a}+c\) | \(\displaystyle\int e^{5x-2}\,\mathrm{d}x=\dfrac{e^{5x-2}}{5}+c\) |
Reciprocal integrals: the power \(-1\)
| Integral | Result | Example |
|---|---|---|
| \(\displaystyle\int\dfrac1x\,\mathrm{d}x\) | \(\ln x+c,\ x>0\) | \(\displaystyle\int\dfrac3x\,\mathrm{d}x=3\ln x+c\) |
| \(\displaystyle\int\dfrac{1}{ax+b}\,\mathrm{d}x\) | \(\dfrac{\ln(ax+b)}{a}+c,\ ax+b>0\) | \(\displaystyle\int\dfrac{1}{4x+3}\,\mathrm{d}x=\dfrac{\ln(4x+3)}{4}+c\) |
Logarithm or power rule? Check the index
| Integrand | Working |
|---|---|
| \(\dfrac{1}{4x+3}=(4x+3)^{-1}\) | \(\displaystyle\int(4x+3)^{-1}\,\mathrm{d}x=\dfrac{\ln(4x+3)}{4}+c\) |
| \(\dfrac{1}{(4x+3)^2}=(4x+3)^{-2}\) | \(\displaystyle\int(4x+3)^{-2}\,\mathrm{d}x=\dfrac{(4x+3)^{-1}}{(4)(-1)}+c\) \(=-\dfrac{1}{4(4x+3)}+c\) |
Only the power \(-1\) produces a logarithm. For any other power of \(ax+b\), stay on the power rule and divide by \(a(n+1)\).
In a formula containing division by \(a\), assume \(a\ne0\). If \(a=0\), integrate the resulting constant separately, provided it is defined. Work only on intervals where the original integrand is real and defined; use \(\ln|ax+b|\) for a reciprocal linear integrand.
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