Antiderivatives of exponential and trigonometric functions (TCE Mathematics Methods, Tasmania)

Syllabus dot point

“Find antiderivatives of exponential and trigonometric functions and apply them in definite integrals.”

8 min readUpdated 2 Jun 2026

Quick answer

Reversing the derivatives gives the standard antiderivatives $\displaystyle\int e^{kx}\,dx = \tfrac{1}{k}e^{kx}+C$ , $\displaystyle\int \sin(kx)\,dx = -\tfrac{1}{k}\cos(kx)+C$ and $\displaystyle\int \cos(kx)\,dx = \tfrac{1}{k}\sin(kx)+C$ . The reciprocal factor $\tfrac{1}{k}$ undoes the chain-rule factor that differentiation produced.

What this dot point is asking

Integration is antidifferentiation, so every derivative rule you learned in Unit 3 reverses into an integration rule. This dot point collects the antiderivatives of the exponential and trigonometric functions, completing the toolkit alongside the power and reciprocal rules.

Reversing each derivative

Because $\dfrac{d}{dx}(e^{kx}) = ke^{kx}$ , integrating $e^{kx}$ must divide by $k$ to cancel that factor. The trigonometric antiderivatives follow the same logic, with the added care that integrating sine introduces a minus sign.

Standard antiderivatives

$\displaystyle\int e^{kx}\,dx = \frac{1}{k}e^{kx} + C$

$\displaystyle\int \sin(kx)\,dx = -\frac{1}{k}\cos(kx) + C$

$\displaystyle\int \cos(kx)\,dx = \frac{1}{k}\sin(kx) + C$

Integration of sine and cosine swap roles with a sign

Differentiating, $\sin$ leads to $\cos$ and $\cos$ leads to $-\sin$ . Integrating reverses this: the integral of $\cos$ is $+\sin$ , and the integral of $\sin$ is $-\cos$ . The minus sign lands on the antiderivative of sine, the opposite place from where it sat in differentiation.

Checking by differentiating back

The reliable way to confirm any antiderivative is to differentiate your answer and check you recover the original integrand. For instance, differentiating $-\tfrac{1}{3}\cos(3x)$ gives $-\tfrac{1}{3}\cdot(-3\sin(3x)) = \sin(3x)$ , which confirms the antiderivative of $\sin(3x)$ .

Worked example

A definite integral with exponential and trigonometric terms

Evaluate $\displaystyle\int_0^{\pi/2} \big(e^{2x} + \cos x\big)\,dx$ .

Step 1: Find the antiderivative of each term

Because the integrand is a sum, we antidifferentiate each term separately. For $e^{2x}$ the coefficient of $x$ is $2$ , so the antiderivative carries a reciprocal factor of $\tfrac{1}{2}$ . For $\cos x$ the antiderivative is $\sin x$ :

\int \big(e^{2x} + \cos x\big)\,dx = \frac{1}{2}e^{2x} + \sin x + C.

Step 2: Apply the fundamental theorem by substituting the limits

Evaluate the antiderivative at the upper limit $x = \tfrac{\pi}{2}$ and subtract its value at the lower limit $x = 0$ :

\left[\frac{1}{2}e^{2x} + \sin x\right]_0^{\pi/2} = \left(\frac{1}{2}e^{\pi} + \sin\frac{\pi}{2}\right) - \left(\frac{1}{2}e^{0} + \sin 0\right).

Step 3: Simplify using known values

Substitute the exact values $\sin\tfrac{\pi}{2} = 1$ , $e^0 = 1$ and $\sin 0 = 0$ :

\left(\frac{1}{2}e^{\pi} + 1\right) - \left(\frac{1}{2} + 0\right) = \frac{1}{2}e^{\pi} + \frac{1}{2}.

Numerically $e^{\pi} \approx 23.14$ , so the integral is about $\tfrac{1}{2}(23.14) + 0.5 \approx 12.07$ .

Final answer: $\dfrac{1}{2}e^{\pi} + \dfrac{1}{2}$ (approximately $12.07$ ).

Linear inner functions

Each rule extends to a linear inner function $kx + c$ , because the chain rule only contributes the constant factor $k$ . So

\int e^{kx + c}\,dx = \frac{1}{k}e^{kx + c} + C, \qquad \int \sin(kx + c)\,dx = -\frac{1}{k}\cos(kx + c) + C,

and similarly for cosine. The shift

c

does not change the reciprocal factor; only the coefficient of

x

matters. For example,

\displaystyle\int \cos(2x - 1)\,dx = \frac{1}{2}\sin(2x - 1) + C

. This linear-substitution shortcut covers the large majority of TASC antidifferentiation questions, because the inner functions in the course are almost always linear.

A definite integral with a phase shift

Evaluate $\displaystyle\int_{0}^{\pi/4} \sin\!\left(2x + \tfrac{\pi}{2}\right)\,dx$ .

Step 1: Find the antiderivative of the shifted sine

The inner function is $2x + \tfrac{\pi}{2}$ , a linear expression with coefficient $k = 2$ on $x$ . The antiderivative of $\sin(kx + c)$ is $-\tfrac{1}{k}\cos(kx + c)$ ; the phase shift $\tfrac{\pi}{2}$ travels through unchanged into the antiderivative:

\int \sin\!\left(2x + \tfrac{\pi}{2}\right)\,dx = -\frac{1}{2}\cos\!\left(2x + \tfrac{\pi}{2}\right) + C.

Step 2: Apply the limits and evaluate

Substitute $x = \tfrac{\pi}{4}$ and $x = 0$ into the antiderivative and subtract. At $x = \tfrac{\pi}{4}$ the argument becomes $\tfrac{\pi}{2} + \tfrac{\pi}{2} = \pi$ ; at $x = 0$ it becomes $\tfrac{\pi}{2}$ :

\left[-\frac{1}{2}\cos\!\left(2x + \tfrac{\pi}{2}\right)\right]_{0}^{\pi/4} = -\frac{1}{2}\cos(\pi) + \frac{1}{2}\cos\!\left(\tfrac{\pi}{2}\right) = -\frac{1}{2}(-1) + \frac{1}{2}(0) = \frac{1}{2}.

The phase shift $\dfrac{\pi}{2}$ rides along untouched; only the factor $\dfrac{1}{2}$ from $k = 2$ enters the antiderivative.

Final answer: $\dfrac{1}{2}$ .

Where these antiderivatives are used

These rules let you find areas under exponential and trigonometric curves, recover totals from exponential or oscillating rates of change, and compute the mean of a continuous random variable whose density involves $e^{-kx}$ . They also combine with the area-between-curves method whenever one of the boundaries is an exponential or trigonometric graph.

Forgetting the reciprocal factor or misplacing the sign

Two slips dominate. Writing $\displaystyle\int e^{3x}\,dx = e^{3x}+C$ omits the $\tfrac{1}{3}$ factor; the correct result is $\tfrac{1}{3}e^{3x}+C$ . Writing $\displaystyle\int \sin x\,dx = \cos x + C$ drops the minus; the correct result is $-\cos x + C$ . Differentiating your answer to recover the integrand catches both errors immediately.

Summary

Reverse each derivative: dividing by $k$ for $e^{kx}$ and for $\sin(kx)$ or $\cos(kx)$ , and remembering that the integral of sine carries a minus sign while the integral of cosine does not. Always verify by differentiating back. With these antiderivatives added to the power and reciprocal rules, you can integrate any standard function in the course and apply it in definite integrals.

Exam-style questions

Questions in the style of TASC exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.

Practice4 marks

Calculator-free. Evaluate the following integrals: (a)

\displaystyle\int -\cos\!\left(\tfrac{\pi}{3}x\right)\,dx

; (b)

\displaystyle\int_{1/2}^{1} e^{2x - 1}\,dx

Show worked answer →

(a) The antiderivative of $\cos(kx)$ is $\dfrac{1}{k}\sin(kx)$ . Here $k = \dfrac{\pi}{3}$ , so $\dfrac{1}{k} = \dfrac{3}{\pi}$ :

\int -\cos\!\left(\tfrac{\pi}{3}x\right)\,dx = -\frac{3}{\pi}\sin\!\left(\tfrac{\pi}{3}x\right) + C.

The reciprocal factor

\dfrac{3}{\pi}

comes from reversing the chain rule.

(b) The antiderivative of $e^{2x - 1}$ is $\dfrac{1}{2}e^{2x - 1}$ (divide by the coefficient of $x$ ).

\left[\tfrac{1}{2}e^{2x - 1}\right]_{1/2}^{1} = \tfrac{1}{2}e^{1} - \tfrac{1}{2}e^{0} = \tfrac{1}{2}(e - 1).

Marks reward the correct reciprocal factor in each case and the correct substitution of limits in (b).

Practice4 marks

Calculator-assumed. (a) Evaluate

\displaystyle\int \sqrt{2x - 3}\,dx

. (b) Evaluate

\displaystyle\int \dfrac{e^{5x + 1}}{e^{2x - 3}}\,dx

Show worked answer →

(a) Write $\sqrt{2x - 3} = (2x - 3)^{1/2}$ . Antidifferentiate using the rule for a linear inner function: raise the power, divide by the new power, then divide by the coefficient of $x$ (which is $2$ ):

\int (2x - 3)^{1/2}\,dx = \frac{1}{2}\cdot\frac{2}{3}(2x - 3)^{3/2} + C = \frac{1}{3}(2x - 3)^{3/2} + C.

(b) First simplify using index laws: $\dfrac{e^{5x + 1}}{e^{2x - 3}} = e^{(5x + 1) - (2x - 3)} = e^{3x + 4}$ .

\int e^{3x + 4}\,dx = \frac{1}{3}e^{3x + 4} + C.

The key step in (b) is simplifying to a single exponential before integrating; one mark for that, one for the antiderivative.

Practise this

Sources & how we know this

Mathematics Methods Level 4 (MTM415117) course document (2017 onwards) — TASC