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How do we integrate exponential, reciprocal and trigonometric functions, including those of a linear expression?

Integrate the exponential, reciprocal and trigonometric functions and their linear composites as the reverse of the corresponding derivatives

A focused answer to the O-Level A-Maths outcome on integrating standard functions. The integrals of the exponential, reciprocal and trigonometric functions, and the rule for a linear composite.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to integrate the standard functions beyond powers: the exponential exe^x, the reciprocal 1x\dfrac{1}{x}, and the trigonometric functions sinx\sin x and cosx\cos x, together with their linear composites such as e2xe^{2x} or cos3x\cos 3x. Each is the reverse of a derivative you already know.

The answer

The standard integrals

Reverse the standard derivatives:

exdx=ex+C,1xdx=lnx+C,\int e^x\,dx = e^x + C, \qquad \int \frac{1}{x}\,dx = \ln|x| + C,

sinxdx=cosx+C,cosxdx=sinx+C.\int \sin x\,dx = -\cos x + C, \qquad \int \cos x\,dx = \sin x + C.

Note the minus sign when integrating sine, and the modulus in the logarithm.

Why the reciprocal gives a logarithm

The power rule fails for x1x^{-1} because it would divide by zero. Instead, since ddxlnx=1x\dfrac{d}{dx}\ln x = \dfrac{1}{x}, the integral of 1x\dfrac{1}{x} is lnx\ln|x|. The modulus allows for negative xx.

Linear composites

For a function of (ax+b)(ax + b), integrate as the simple version and divide by the coefficient aa of xx inside, just as with powers:

eax+bdx=1aeax+b+C,cos(ax+b)dx=1asin(ax+b)+C.\int e^{ax + b}\,dx = \frac{1}{a}e^{ax + b} + C, \qquad \int \cos(ax + b)\,dx = \frac{1}{a}\sin(ax + b) + C.

The division by aa undoes the chain-rule factor differentiation would have introduced.

Putting it together

Integrate a sum term by term, applying the right standard integral and the inner-coefficient division to each. Keep one constant of integration for the whole expression.

Definite integrals of these functions

The same standard results apply to definite integrals: integrate first, then substitute the upper and lower limits and subtract. For exponentials and trigonometric functions this often produces neat exact values, such as 0πsinxdx=[cosx]0π=2\int_0^{\pi} \sin x\,dx = [-\cos x]_0^{\pi} = 2, which examiners expect to be left exact rather than rounded.

Checking by differentiating

Since integration reverses differentiation, differentiate your answer to confirm it returns the integrand. This quickly exposes a dropped minus sign on a sine, a missing modulus on a logarithm, or a forgotten inner-coefficient division on a composite.

Examples in context

Example 1. Charge from a current. An electric current that varies as I=I0et/τI = I_0 e^{-t/\tau} integrates to give the total charge that has flowed, the exponential integral underpinning capacitor discharge.

Example 2. Displacement of an oscillator. A velocity varying as v=cos(ωt)v = \cos(\omega t) integrates to a sine displacement, so integrating trigonometric functions describes how an oscillating object moves over time.

Try this

Q1. Find e3xdx\displaystyle\int e^{3x}\,dx. [2 marks]

  • Cue. 13e3x+C\dfrac{1}{3}e^{3x} + C.

Q2. Find cos2xdx\displaystyle\int \cos 2x\,dx. [2 marks]

  • Cue. 12sin2x+C\dfrac{1}{2}\sin 2x + C.

Q3. Find 5xdx\displaystyle\int \dfrac{5}{x}\,dx. [2 marks]

  • Cue. 5lnx+C5\ln|x| + C.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original4 marksFind (e2x+3x)dx\displaystyle\int \left(e^{2x} + \dfrac{3}{x}\right) dx.
Show worked answer →

Integrate the exponential of a linear expression: e2xdx=12e2x\displaystyle\int e^{2x}\,dx = \dfrac{1}{2}e^{2x} (divide by the coefficient of xx).

Integrate the reciprocal: 3xdx=3lnx\displaystyle\int \dfrac{3}{x}\,dx = 3\ln|x|.

So the integral is 12e2x+3lnx+C\dfrac{1}{2}e^{2x} + 3\ln|x| + C.

Markers reward the exponential with the division by 22, the logarithm with the modulus, and the constant CC.

Original4 marksFind (cos3xsinx)dx\displaystyle\int \big(\cos 3x - \sin x\big)\,dx.
Show worked answer →

Reverse the trigonometric derivatives. cos3xdx=13sin3x\displaystyle\int \cos 3x\,dx = \dfrac{1}{3}\sin 3x (divide by the inner coefficient 33).

sinxdx=cosx\displaystyle\int -\sin x\,dx = \cos x.

So the integral is 13sin3x+cosx+C\dfrac{1}{3}\sin 3x + \cos x + C.

Markers reward cos3xdx=13sin3x\int\cos 3x\,dx = \tfrac{1}{3}\sin 3x, sinxdx=cosx\int\sin x\,dx = -\cos x handled with its sign, and the constant.

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