1.08c - Integrating exponential, logarithmic and trigonometric forms | OCR A-Level Mathematics A

Integrate exponential, reciprocal, sine and cosine forms, controlling signs, input multipliers and logarithmic domains. Rewrite trigonometric powers with identities and use a supplied inverse-function integral correctly.

Reverse exponential derivatives

Do $\sin^2x$ and $\cos^2x$ enclose the same area over $0\le x\le\pi$ ? At zero their heights are $0$ and $1$ , so they certainly do not match point by point. We will find an exact answer by expanding our collection of antiderivatives, starting with exponentials.

You already know

An indefinite integral includes an arbitrary constant. Differentiate an answer to check it: $\mathrm d(\mathrm e^{kx})/\mathrm dx=k\mathrm e^{kx}$ , and calculus with trigonometric functions uses radians.

Integration must undo the multiplier introduced by a scaled input.

Exponential integral

For a constant $k\ne0$ ,

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

If $k=0$ , the integrand is $1$ , so integrate directly to $x+C$ .

Not in the formula sheet: learn it.

Handle a sum one term at a time; the exponential's coefficient and its input multiplier have different jobs.

Worked example 1: growth and decay together

Question. Find $\int(3\mathrm e^{2x}-4\mathrm e^{-x}+2)\,\mathrm dx$ .

Plan. Divide each exponential coefficient by its input multiplier, then integrate the constant.

\begin{aligned} \int3\mathrm e^{2x}\,\mathrm dx&=\frac32\mathrm e^{2x} &&\text{divide by }2\\ \int-4\mathrm e^{-x}\,\mathrm dx&=4\mathrm e^{-x} &&\text{divide by }-1\\ \int(3\mathrm e^{2x}-4\mathrm e^{-x}+2)\,\mathrm dx &=\frac32\mathrm e^{2x}+4\mathrm e^{-x}+2x+C \end{aligned}

Why this step? The negative input multiplier cancels the negative coefficient in the second integral.

Answer. $\tfrac32\mathrm e^{2x}+4\mathrm e^{-x}+2x+C$ .

Check. Differentiating the three nonconstant terms gives $3\mathrm e^{2x}-4\mathrm e^{-x}+2$ .

Change only the sign of the second exponential's input, from $-x$ to $x$ .

Worked example 2: the sign comes from the input

Question. Find $\int(3\mathrm e^{2x}-4\mathrm e^x+2)\,\mathrm dx$ .

Plan. Keep the first and last integrals and use multiplier $1$ for the middle term.

\begin{aligned} \int-4\mathrm e^x\,\mathrm dx&=-4\mathrm e^x &&\text{divide by }1\\ \int(3\mathrm e^{2x}-4\mathrm e^x+2)\,\mathrm dx &=\frac32\mathrm e^{2x}-4\mathrm e^x+2x+C \end{aligned}

Why this step? A negative coefficient alone does not force an integral to become positive; the input derivative decides the extra factor.

Answer. $\tfrac32\mathrm e^{2x}-4\mathrm e^x+2x+C$ .

Check. The derivative of $-4\mathrm e^x$ is $-4\mathrm e^x$ .

One constant represents the entire sum. Carrying a separate arbitrary constant for each term adds no new information.

Watch out

Wrong: $\int\mathrm e^{2x}\,\mathrm dx=2\mathrm e^{2x}+C$ . Its derivative is $4\mathrm e^{2x}$ . Correct: $\tfrac12\mathrm e^{2x}+C$ .

Now change only the second input multiplier again.

The reciprocal integral needs an absolute value

The power rule stops at exponent $-1$ . The logarithmic derivative supplies the missing integral.

Reciprocal integral

On an interval not containing zero,

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

For $x>0$ this is $\ln x+C$ ; for $x<0$ it is $\ln(-x)+C$ . Indeed, differentiating $\ln(-x)$ gives $(-1)/(-x)=1/x$ . The sheet gives the more general logarithmic-derivative integral; this standard reciprocal result is required recall.

Not in the formula sheet: learn it.

The absolute value makes the logarithm's argument positive. It does not define the integrand at zero, and constants on the positive and negative branches may be chosen separately.

Worked example 3: a positive interval

Question. Find $\int(4/x+2x)\,\mathrm dx$ for $x>0$ .

Plan. Use the logarithmic rule for the reciprocal and the power rule for $2x$ .

\begin{aligned} \int\left(\frac4x+2x\right)\,\mathrm dx &=4\ln x+x^2+C &&\text{integrate term by term}\\ \frac{\mathrm d}{\mathrm dx}(4\ln x+x^2+C)&=\frac4x+2x &&\text{check} \end{aligned}

Why this step? The reciprocal is not a power-rule case; treating it separately avoids division by zero.

Answer. $4\ln x+x^2+C$ , for $x>0$ .

Check. $\ln x$ is real throughout this stated interval.

Change only the domain, to the negative branch. The integrand and its algebraic coefficients stay unchanged.

Worked example 4: a negative interval

Question. Find $\int(4/x+2x)\,\mathrm dx$ for $x<0$ .

Plan. Use $\ln(-x)$ so the logarithm is real on the stated branch.

\begin{aligned} \int\left(\frac4x+2x\right)\,\mathrm dx&=4\ln(-x)+x^2+C\\ \frac{\mathrm d}{\mathrm dx}(4\ln(-x)+x^2+C)&=4\frac{-1}{-x}+2x=\frac4x+2x \end{aligned}

Why this step? For negative $x$ , $-x>0$ , whereas $\ln x$ is not real.

Answer. $4\ln(-x)+x^2+C$ , equivalently $4\ln|x|+x^2+C$ , on $x<0$ .

Check. At $x=-1$ , the derivative is $-4-2=-6$ , matching the integrand.

The formula works on either branch, but it cannot be used to evaluate an ordinary integral through zero.

Watch out

Integrating $1/x$ gives a logarithm. Integrating $\ln x$ is a different task; its answer is not $1/x$ . Differentiate to check the direction of the operation.

Keep the negative interval and change only the reciprocal coefficient.

Trigonometric integrals use radians

The minus sign in a sine integral comes from the derivative of cosine. A second factor reverses any scaling of the input.

Sine and cosine integrals

For a constant $k\ne0$ , with arguments in radians,

\int\sin(kx)\,\mathrm dx=-\frac{\cos(kx)}k+C,\qquad \int\cos(kx)\,\mathrm dx=\frac{\sin(kx)}k+C

At $k=0$ , integrate the constant functions $0$ and $1$ directly.

Not in the formula sheet: learn it.

Check both sign and multiplier by differentiation, rather than relying on the appearance of a remembered formula.

Worked example 5: two scaled inputs

Question. Find $\int(6\sin3x-4\cos2x)\,\mathrm dx$ , with $x$ in radians.

Plan. Integrate each term, dividing by its own input multiplier.

\begin{aligned} \int6\sin3x\,\mathrm dx&=-2\cos3x &&\text{minus cosine, divide by }3\\ \int-4\cos2x\,\mathrm dx&=-2\sin2x &&\text{divide by }2\\ \int(6\sin3x-4\cos2x)\,\mathrm dx&=-2\cos3x-2\sin2x+C \end{aligned}

Why this step? Each term has its own input multiplier; there is no single factor to apply to the whole sum.

Answer. $-2\cos3x-2\sin2x+C$ .

Check. Differentiation gives $6\sin3x-4\cos2x$ .

Change only the sign of the sine coefficient. The cosine term is unchanged.

Continue this lesson.

You’re previewing NovaLearn without an account. Create a free account to read the rest.

Create free account

Last Accessed

Pure Mathematics

Statistics

Mechanics

1.08c - Integrating exponential, logarithmic and trigonometric forms

Reverse exponential derivatives

The reciprocal integral needs an absolute value

Trigonometric integrals use radians

Continue this lesson.

Learn faster with Nova