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Pure · Integration

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Integrating standard functions

The integrals that are just the derivative rules in reverse — e^x, 1/x → ln|x|, sin, cos and sec² — together with the f(ax+b) rule for a linear inside function (integrate as normal, then divide by a).

Pure · Integration

Integrating standard functions

The integrals that are just the derivative rules in reverse — e^x, 1/x → ln|x|, sin, cos and sec² — together with the f(ax+b) rule for a linear inside function (integrate as normal, then divide by a).

Why it works

Integration is differentiation run backwards, so every derivative you know gives you an integral for free — just read the rule the other way. Once you have the Year 2 derivatives of exe^x, ln⁡x\ln x and the trig functions, a whole table of standard integrals falls out: ∫ex dx=ex+c,∫cos⁡x dx=sin⁡x+c,∫sin⁡x dx=−cos⁡x+c,\int e^x\,dx = e^x + c, \qquad \int \cos x\,dx = \sin x + c, \qquad \int \sin x\,dx = -\cos x + c, ∫sec⁡2x dx=tan⁡x+c,∫sec⁡xtan⁡x dx=sec⁡x+c,∫cosec⁡2x dx=−cot⁡x+c.\int \sec^2 x\,dx = \tan x + c, \qquad \int \sec x\tan x\,dx = \sec x + c, \qquad \int \operatorname{cosec}^2 x\,dx = -\cot x + c.

The special one: 1/x\boldsymbol{1/x}. The power rule for integration, "raise the power and divide by the new power", dies for ∫x−1 dx\int x^{-1}\,dx — it would divide by zero. But you already know a function whose derivative is 1x\tfrac1x: it is ln⁡x\ln x. So ∫1x dx=ln⁡∣x∣+c.\int \frac{1}{x}\,dx = \ln|x| + c. The modulus matters: ln⁡x\ln x only exists for x>0x > 0, yet 1x\tfrac1x is perfectly fine for x<0x < 0 too. Writing ln⁡∣x∣\ln|x| covers both halves (for x<0x<0, ddxln⁡(−x)=−1−x=1x\frac{d}{dx}\ln(-x) = \frac{-1}{-x} = \frac1x). Always include the modulus when you integrate a reciprocal.

The f(ax+b)\boldsymbol{f(ax+b)} rule — a linear inside function. What about ∫e3x dx\int e^{3x}\,dx or ∫cos⁡2x dx\int \cos 2x\,dx? Differentiating F(ax+b)F(ax+b) brings out an extra factor aa by the chain rule, so to undo that you must divide by aa: ∫f(ax+b) dx=1a F(ax+b)+c,where F′=f.\int f(ax + b)\,dx = \frac{1}{a}\,F(ax + b) + c, \quad\text{where } F' = f. So ∫e3x dx=13e3x+c\int e^{3x}\,dx = \tfrac13 e^{3x} + c, ∫cos⁡2x dx=12sin⁡2x+c\int \cos 2x\,dx = \tfrac12\sin 2x + c, and ∫(2x+1)5 dx=12⋅16(2x+1)6+c=112(2x+1)6+c\int (2x+1)^5\,dx = \tfrac{1}{2}\cdot\tfrac{1}{6}(2x+1)^6 + c = \tfrac{1}{12}(2x+1)^6 + c. This shortcut works only because the inside is linear (ax+bax + b); a non-linear inside needs substitution or the reverse chain rule instead.

A reciprocal of a linear function combines both ideas: ∫1ax+b dx=1aln⁡∣ax+b∣+c.\int \frac{1}{ax + b}\,dx = \frac{1}{a}\ln|ax + b| + c.