Mc24 - Integrating exponential, logarithmic and trigonometric forms | OCR A-Level Mathematics B (MEI)

Integrate exponential, reciprocal, sine and cosine functions and their sums. Reverse input multipliers accurately, handle logarithms on real intervals and use identities to turn trigonometric powers into standard forms.

Reverse exponential derivatives

Does tripling the number of waves change the area under a squared sine curve? At $x=\pi/6$ , $\sin^2x=1/4$ , but $\sin^2(3x)=1$ . Yet we will show that both enclose the same area on $[0,\pi]$ . First we need some standard integrals and a reliable way to check their factors.

You already know

Differentiate to check an antiderivative. In particular, $\mathrm d(\mathrm e^{kx})/\mathrm dx=k\mathrm e^{kx}$ ; an indefinite integral includes an arbitrary constant. Trigonometric calculus uses radians.

Integrating a scaled input reverses its derivative's multiplier.

Integrating an exponential

For a constant $k\ne0$ ,

\int\mathrm e^{kx}\,\mathrm dx=\frac1k\mathrm e^{kx}+c

At $k=0$ , integrate $1$ directly to obtain $x+c$ .

Not in the formula sheet: learn it.

The input multiplier may be negative. Let the division determine the sign.

Worked example 1: a decaying exponential

Question. Find $\int(5\mathrm e^{-2x}+3)\,\mathrm dx$ .

Plan. Divide the exponential coefficient by $-2$ , then integrate the constant.

\begin{aligned} \int5\mathrm e^{-2x}\,\mathrm dx&=-\frac52\mathrm e^{-2x} &&\text{reverse the input factor}\\ \int(5\mathrm e^{-2x}+3)\,\mathrm dx&=-\frac52\mathrm e^{-2x}+3x+c \end{aligned}

Why this step? Differentiating a negative exponential introduces a negative factor, so its antiderivative coefficient must compensate.

Answer. $-\tfrac52\mathrm e^{-2x}+3x+c$ .

Check. Differentiating gives $(-5/2)(-2)\mathrm e^{-2x}+3=5\mathrm e^{-2x}+3$ .

Change only the sign of the exponential's input multiplier.

Worked example 2: the corresponding growing exponential

Question. Find $\int(5\mathrm e^{2x}+3)\,\mathrm dx$ .

Plan. Divide by $2$ and keep the integral of the constant term.

\begin{aligned} \int5\mathrm e^{2x}\,\mathrm dx&=\frac52\mathrm e^{2x} &&\text{positive input factor}\\ \int(5\mathrm e^{2x}+3)\,\mathrm dx&=\frac52\mathrm e^{2x}+3x+c \end{aligned}

Why this step? Changing the sign inside the exponential changes the sign of the compensating coefficient.

Answer. $\tfrac52\mathrm e^{2x}+3x+c$ .

Check. Its derivative is $5\mathrm e^{2x}+3$ .

For a longer sum, apply the same rule separately to every term. For example, $3\mathrm e^{2x}-4\mathrm e^{-x}+2$ has antiderivative $\tfrac32\mathrm e^{2x}+4\mathrm e^{-x}+2x+c$ ; differentiating recovers all three terms. One constant covers the whole sum.

Watch out

Wrong: $\int\mathrm e^{2x}\,\mathrm dx=2\mathrm e^{2x}+c$ . This differentiates to $4\mathrm e^{2x}$ . Divide by $2$ instead of multiplying.

Change only the input multiplier's magnitude.

The reciprocal integral needs an absolute value

The reciprocal $1/x$ is the exception to the power rule. Its antiderivative comes from a logarithm, with a domain decision built in.

The reciprocal integral

On any interval avoiding zero,

\int\frac1x\,\mathrm dx=\ln|x|+c

For $x>0$ , use $\ln x$ ; for $x<0$ , use $\ln(-x)$ . The latter has derivative $(-1)/(-x)=1/x$ . The sheet prints the general logarithmic-derivative integral; this standard reciprocal form is required recall.

Not in the formula sheet: learn it.

The modulus bars make the logarithm real on either branch. They do not remove the undefined point at zero, and the integration constants on the two separate branches need not agree.

Worked example 3: the positive branch

Question. Find $\int(3/x-2)\,\mathrm dx$ , for $x>0$ .

Plan. Integrate the reciprocal logarithmically and the constant as a linear term.

\begin{aligned} \int\left(\frac3x-2\right)\,\mathrm dx&=3\ln x-2x+c\\ \frac{\mathrm d}{\mathrm dx}(3\ln x-2x+c)&=\frac3x-2 &&\text{check the derivative} \end{aligned}

Why this step? Exponent $-1$ would give a zero denominator in the power rule; a logarithm supplies the correct derivative.

Answer. $3\ln x-2x+c$ , on $x>0$ .

Check. At $x=1$ , the derivative is $1$ , matching $3/1-2$ .

Change only the interval to $x<0$ . The expression being integrated is unchanged.

Worked example 4: the negative branch

Question. Find $\int(3/x-2)\,\mathrm dx$ , for $x<0$ .

Plan. Use $\ln(-x)$ to keep the logarithm's argument positive.

\begin{aligned} \int\left(\frac3x-2\right)\,\mathrm dx&=3\ln(-x)-2x+c\\ \frac{\mathrm d}{\mathrm dx}(3\ln(-x)-2x+c)&=3\frac{-1}{-x}-2=\frac3x-2 \end{aligned}

Why this step? A real formula must work on the specified interval; $\ln x$ is not real there.

Answer. $3\ln(-x)-2x+c$ , equivalently $3\ln|x|-2x+c$ , for $x<0$ .

Check. At $x=-1$ , the derivative is $-5$ , matching the integrand.

A definite interval must avoid zero too; two finite endpoint values do not legitimise an integral through a singularity.

Watch out

The operations have opposite directions: integrating $1/x$ gives $\ln|x|+c$ , whereas differentiating $\ln x$ gives $1/x$ . Integrating $\ln x$ is a separate problem, using a later method.

Keep the negative branch and vary only the reciprocal coefficient.

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Mc24 - Integrating exponential, logarithmic and trigonometric forms

Reverse exponential derivatives

The reciprocal integral needs an absolute value

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