Integrating exponentials and trigonometric functionsAQA A-Level Maths: Revision notes

Section 1

Integrating exponentials

Reversing differentiation: since $\frac{d}{dx}e^{kx}=ke^{kx}$ , $\int e^{kx}\,dx=\frac1ke^{kx}+c\qquad(k\ne0).$ The constant of integration $c$ must appear in every indefinite integral. A constant multiple stays outside: $\int6e^{3x}dx=6\times\frac13e^{3x}+c=2e^{3x}+c$ . The base $e$ is special because $e^{kx}$ keeps its form when differentiated or integrated; do not use the power rule on the index.

Key termsindefinite integralconstant of integration

Common mistake

Multiplying by $k$ instead of dividing: $\int e^{3x}dx=\frac13e^{3x}$ , not $3e^{3x}$ .

Exam tip

Differentiate your answer to check it returns the original function.

Section 2

Integrating 1/x

$\int\frac1x\,dx=\ln|x|+c\qquad(x\ne0).$ This is the case the power rule cannot do ( $x^{-1}$ would need division by 0). A linear inside works like the exponential: $\int\frac{1}{kx}dx=\frac1k\ln|x|+c$ . For $x>0$ you may write $\ln x$ . In definite integrals use the laws of logarithms to simplify: $\int_\pi^{2\pi}\frac3x\,dx=3(\ln2\pi-\ln\pi)=3\ln2$ .

Key termsln|x|

Common mistake

Writing $\int\frac1x dx=\frac{x^0}{0}$ ; the power rule fails for the index $-1$ .

Section 3

Integrating sin kx and cos kx

With $x$ in radians: $\int\sin kx\,dx=-\frac1k\cos kx+c,\qquad\int\cos kx\,dx=\frac1k\sin kx+c.$ These come from $\frac{d}{dx}\cos kx=-k\sin kx$ and $\frac{d}{dx}\sin kx=k\cos kx$ . So $\int4\cos2t\,dt=2\sin2t+c$ and $\int5\cos\frac x2\,dx=10\sin\frac x2+c$ (dividing by $\frac12$ is multiplying by 2).

Key termsradians

Common mistake

Losing the minus sign: $\int\sin kx\,dx$ is negative, $\int\cos kx\,dx$ is positive.

Exam tip

In definite integrals check the calculator is in radian mode.

Section 4

Sums, differences and constant multiples

Integration is linear: $\int\left[af(x)\pm bg(x)\right]dx=a\int f\,dx\pm b\int g\,dx$ . Integrate each term separately and include a single constant: $\int\left(\frac3x-4e^{-2x}+5\cos\frac x2\right)dx=3\ln x+2e^{-2x}+10\sin\frac x2+c.$ For a definite integral substitute the limits into the whole antiderivative (no $c$ needed) and subtract: $\int_0^{\ln2}6e^{3x}dx=\left[2e^{3x}\right]_0^{\ln2}=2(8)-2=14$ , using $e^{3\ln2}=2^3$ .

Key termslinearity

Exam tip

Use $e^{k\ln a}=a^k$ and $e^0=1$ to simplify exponentials at the limits.

Section 5

Finding constants and interpreting results

A point on the curve fixes $c$ : if $\frac{dy}{dx}=6e^{3x}$ through $(0,5)$ then $y=2e^{3x}+c$ and $2+c=5$ , so $y=2e^{3x}+3$ . In kinematics integrating velocity gives displacement, not distance. If $v$ changes sign, split the integral at the turning point and add the sizes. A model such as $R=20e^{-0.1t}+6\sin\frac{\pi t}{30}$ may give impossible values (a negative inflow) outside a limited time, so comment on validity.

Key termsdisplacementdistance

Common mistake

Giving the displacement ( $0$ ) when asked for the total distance after a turning point.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Integrating exponentials and trigonometric functions

A curve has gradient function $\frac{dy}{dx}=6e^{3x}$ and passes through the point $(0,5)$ .
Find the exact value of $\int_0^{\ln 2}6e^{3x}\,dx$ .2 marks
A particle moves on a straight line. At time $t$ seconds its velocity is $v=4\cos 2t$ m s $^{-1}$ , where $2t$ is in radians. The particle starts at the origin, so its displacement is $s=0$ when $t=0$ .
Find the total distance travelled by the particle between $t=0$ and $t=\frac{\pi}{2}$ .2 marks
Let $f(x)=\frac{3}{x}-4e^{-2x}+5\cos\frac{x}{2}$ and $h(x)=\frac{3}{x}+5\cos\frac{x}{2}$ , for $x>0$ , where angles are in radians.
Find $\int f(x)\,dx$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).