2.5, 3.5 - Exponential and trigonometric integrals | Cambridge A-Level Mathematics

Integration reverses differentiation, but a function inside another function leaves a constant factor to undo. In 2.5 and 3.5, you will use that idea to integrate exponentials, reciprocals of linear expressions and three standard trigonometric forms. Checking by differentiation will keep the factors and signs under control.

Linear exponential integrands

What function differentiates to $e^{3x}$ ? It is tempting to answer $e^{3x}$ , but that differentiates to $3e^{3x}$ . Dividing by $3$ first gives the required derivative:

\frac{\mathrm d}{\mathrm dx}\left(\frac13e^{3x}\right)=\frac13\times3e^{3x}=e^{3x}.

This is reverse chain rule reasoning: predict an antiderivative, differentiate it, then adjust the constant multiplier. The inner expression here is linear, so its derivative is a constant that we can divide out.

Integrating a linear exponential

For constants $a\ne0$ , $b$ and $k$ ,

\int k e^{ax+b}\,\mathrm dx=\frac{k}{a}e^{ax+b}+C.

The basic result $\int e^x\,\mathrm dx=e^x$ , with the arbitrary constant omitted, is in the formula booklet. The linear-inner-function form is not listed separately: reconstruct it by reverse differentiation.

The constant $b$ stays in the exponent. Only $a$ , the coefficient of $x$ , appears in the dividing factor. If $a=0$ , the integrand is constant instead: $\int ke^b\,\mathrm dx=ke^b x+C$ .

Worked example 1: a growing exponential

Question. Find $\int 6e^{3x+1}\,\mathrm dx$ .

Plan. Keep the exponential and divide its coefficient by the inner derivative $3$ .

\int 6e^{3x+1}\,\mathrm dx =\frac63e^{3x+1}+C =2e^{3x+1}+C.

Why divide? Differentiation will multiply the exponential by $3$ , so the coefficient $2$ becomes the required $6$ .

Answer. $2e^{3x+1}+C$ .

Check. $\frac{\mathrm d}{\mathrm dx}(2e^{3x+1}+C)=6e^{3x+1}$ .

Change only the sign of the coefficient of $x$ . The antiderivative now needs a negative multiplier.

Worked example 2: a decaying exponential

Question. Find $\int 6e^{1-3x}\,\mathrm dx$ .

Plan. Divide by the inner derivative $-3$ , retaining its sign.

\int 6e^{1-3x}\,\mathrm dx =\frac6{-3}e^{1-3x}+C =-2e^{1-3x}+C.

Answer. $-2e^{1-3x}+C$ .

Check. Differentiation gives $(-2)(-3)e^{1-3x}=6e^{1-3x}$ , which is positive as required.

For a definite integral, use the same antiderivative and subtract its value at the lower limit from its value at the upper limit. The arbitrary constant cancels. For example, over $0\le x\le1$ , the second integrand has integral $[-2e^{1-3x}]_0^1=2e-2e^{-2}$ , a positive result for a positive function.

Watch out

$\int e^{3x}\,\mathrm dx\ne3e^{3x}+C$ . The chain-rule factor is undone by division, giving $\frac13e^{3x}+C$ . A derivative check detects the error immediately.

Try another negative inner derivative.

Reciprocal linear integrands

The power rule excludes exponent $-1$ . Increasing that exponent would give zero, leaving a division by zero. The missing antiderivative is logarithmic: the derivative of $\ln x$ is $1/x$ for $x>0$ .

For $x<0$ , $\ln(-x)$ is real and has derivative $(-1)/(-x)=1/x$ . These two cases are compactly written as $\ln|x|$ . The absolute value makes the logarithm's argument positive; it does not allow $x=0$ .

Integrating a reciprocal linear expression

For $a\ne0$ , on any interval where $ax+b\ne0$ ,

\int\frac{k}{ax+b}\,\mathrm dx =\frac{k}{a}\ln|ax+b|+C.

The basic result $\int\frac1x\,\mathrm dx=\ln|x|$ , with the arbitrary constant omitted, is in the formula booklet. The linear-inner-function form is not listed separately: reconstruct it by reverse differentiation.

For example, differentiating $\ln|3x-2|$ gives $3/(3x-2)$ . Divide by $3$ to get numerator $1$ . The antiderivative is valid separately on either side of $x=2/3$ ; the arbitrary constants on those two intervals need not be the same.

Worked example 3: a positive logarithm argument

Question. Find $\int\frac6{3x+2}\,\mathrm dx$ , for $x>-2/3$ .

Plan. Use a logarithm and divide the numerator by the derivative of the denominator.

\int\frac6{3x+2}\,\mathrm dx =\frac63\ln(3x+2)+C =2\ln(3x+2)+C.

The stated interval makes $3x+2>0$ , so absolute-value signs are optional here.

Answer. $2\ln(3x+2)+C$ .

Check. Its derivative is $2\times3/(3x+2)=6/(3x+2)$ .

Keep the same integrand but move to the other side of its undefined input. The logarithm must now contain an absolute value or an equivalent positive argument.

Worked example 4: a negative denominator

Question. Find $\int\frac6{3x+2}\,\mathrm dx$ , for $x<-2/3$ .

Plan. Use $\ln|3x+2|$ , since $3x+2$ is negative on this interval.

\int\frac6{3x+2}\,\mathrm dx =2\ln|3x+2|+C =2\ln(-3x-2)+C.

Answer. $2\ln|3x+2|+C$ , equivalently $2\ln(-3x-2)+C$ on the stated interval.

Check. Differentiating the second form gives $2(-3)/(-3x-2)=6/(3x+2)$ .

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2.5, 3.5 - Exponential and trigonometric integrals

Linear exponential integrands

Reciprocal linear integrands

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