Integrating Exponentials and Trigonometric Functions — Edexcel A-level Mathematics (Pure)

SPEC 8.2 • INTEGRATION

Integrating Exponential and Trigonometric Functions

Master the standard integrals for 1/x, eˣ, sin x, cos x, sec², cosec², and more

Reversing Differentiation

Integration reverses differentiation. If we know what differentiates to give a function, we know its integral.

Logarithmic

\frac{d}{d x} (ln x) = \frac{1}{x}

Exponential

\frac{d}{d x} (e^{x}) = e^{x}

\frac{d}{d x} (e^{k x}) = k e^{k x}

Key Insight

To find an integral, ask: "What function differentiates to give this?" Then reverse the process, remembering to divide by any coefficient of x.

Integrating 1/x

Since the derivative of

ln x

is

\frac{1}{x}

, we can reverse this:

Standard Result

\int \frac{1}{x} d x = ln ∣ x ∣ + c

Important: Modulus Signs

We write

ln ∣ x ∣

(with modulus signs) because

ln x

is only defined for positive x, but

\frac{1}{x}

exists for all

x \neq = 0

. The modulus ensures the answer works for negative x too.

Generalisation

\int \frac{1}{a x + b} d x = \frac{1}{a} ln ∣ a x + b ∣ + c

Divide by the coefficient of x (which is a)

Integrating Exponentials

The exponential function

e^{x}

is unique — it is its own derivative and its own integral:

Basic Form

\int e^{x} d x = e^{x} + c

With Coefficient

\int e^{k x} d x = \frac{1}{k} e^{k x} + c

Why Divide by k?

When differentiating

e^{k x}

, the chain rule multiplies by k. So when integrating, we divide by k to reverse this.

FunctionIntegral

e^{2 x}

\frac{1}{2} e^{2 x} + c

e^{- 3 x}

- \frac{1}{3} e^{- 3 x} + c

Worked Example01

Find $\int (\frac{3}{x} + 4 e^{3 x} - 2 e^{- 5 x}) d x$

1

Integrate term by term

2

\int \frac{3}{x} d x = 3 ln ∣ x ∣

3

\int 4 e^{3 x} d x = 4 \times \frac{1}{3} e^{3 x} = \frac{4}{3} e^{3 x}

4

\int - 2 e^{- 5 x} d x = - 2 \times \frac{1}{- 5} e^{- 5 x} = \frac{2}{5} e^{- 5 x}

Answer: $3 ln ∣ x ∣ + \frac{4}{3} e^{3 x} + \frac{2}{5} e^{- 5 x} + c$

Integrating sin, cos, sec²

Standard Results

FunctionIntegralWith (kx)

sin x

- cos x + c

\int sin (k x) d x = - \frac{1}{k} cos (k x) + c

cos x

sin x + c

\int cos (k x) d x = \frac{1}{k} sin (k x) + c

sec^{2} x

tan x + c

\int sec^{2} (k x) d x = \frac{1}{k} tan (k x) + c

Watch the Sign!

When integrating

sin x

, the result is negative cos x. Check by differentiating:

\frac{d}{d x} (- cos x) = sin x

✓

The Rule

When integrating functions of (kx), always divide by k — the coefficient of x.

More Trigonometric Integrals

These results come from reversing differentiation of sec, cosec, and cot:

Standard Results

FunctionIntegralWith (kx)

sec x tan x

sec x + c

\int sec (k x) tan (k x) d x = \frac{1}{k} sec (k x) + c

cosec x cot x

- cosec x + c

\int cosec (k x) cot (k x) d x = - \frac{1}{k} cosec (k x) + c

cosec^{2} x

- cot x + c

\int cosec^{2} (k x) d x = - \frac{1}{k} cot (k x) + c

Why These Work

\frac{d}{d x} (sec x) = sec x tan x

,

\frac{d}{d x} (- cosec x) = cosec x cot x

,

\frac{d}{d x} (- cot x) = cosec^{2} x

Worked Example02

Find $\int (sec x tan x + cosec^{2} x) d x$

1

Integrate term by term using standard results

2

\int sec x tan x d x = sec x

3

\int cosec^{2} x d x = - cot x

4

Combine with + c

Answer: $sec x - cot x + c$

Worked Example03

Find $\int (sin 2 x + cosec^{2} 3 x) d x$

1

Integrate term by term, dividing by coefficient of x

2

\int sin 2 x d x = - \frac{1}{2} cos 2 x

(divide by 2)

3

\int cosec^{2} 3 x d x = - \frac{1}{3} cot 3 x

(divide by 3)

4

Combine with + c

Answer: $- \frac{1}{2} cos 2 x - \frac{1}{3} cot 3 x + c$

Worked Example04

Find $\int (\frac{2}{x} + e^{2 x} + cos 3 x) d x$

1

Integrate term by term

2

\int \frac{2}{x} d x = 2 ln ∣ x ∣

3

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

(divide by 2)

4

\int cos 3 x d x = \frac{1}{3} sin 3 x

(divide by 3)

Answer: $2 ln ∣ x ∣ + \frac{1}{2} e^{2 x} + \frac{1}{3} sin 3 x + c$

Worked Example05

Find $\int (4 sin x - 3 e^{- 2 x} + sec^{2} x) d x$

1

Integrate term by term

2

\int 4 sin x d x = 4 \times (- cos x) = - 4 cos x

3

\int - 3 e^{- 2 x} d x = - 3 \times \frac{1}{- 2} e^{- 2 x} = \frac{3}{2} e^{- 2 x}

4

\int sec^{2} x d x = tan x

Answer: $- 4 cos x + \frac{3}{2} e^{- 2 x} + tan x + c$

Worked Example06

Find $\int (2 sec 3 x tan 3 x - cosec x cot x) d x$

1

Integrate term by term using standard results

2

\int 2 sec 3 x tan 3 x d x = 2 \times \frac{1}{3} sec 3 x = \frac{2}{3} sec 3 x

3

\int - cosec x cot x d x = - (- cosec x) = cosec x

4

Combine with + c

Answer: $\frac{2}{3} sec 3 x + cosec x + c$

Worked Example07

Find $\int (\frac{1}{x} + 2 e^{x} - 5 cos 4 x + sec^{2} 2 x - 3 cosec^{2} x) d x$

1

\int \frac{1}{x} d x = ln ∣ x ∣

2

\int 2 e^{x} d x = 2 e^{x}

3

\int - 5 cos 4 x d x = - 5 \times \frac{1}{4} sin 4 x = - \frac{5}{4} sin 4 x

4

\int sec^{2} 2 x d x = \frac{1}{2} tan 2 x

5

\int - 3 cosec^{2} x d x = - 3 \times (- cot x) = 3 cot x

Answer: $ln ∣ x ∣ + 2 e^{x} - \frac{5}{4} sin 4 x + \frac{1}{2} tan 2 x + 3 cot x + c$

Practice

Question 1: Find

\int (\frac{4}{x} + e^{4 x} - sin x + cosec^{2} x) d x

1

\int \frac{4}{x} d x = 4 ln ∣ x ∣

2

\int e^{4 x} d x = \frac{1}{4} e^{4 x}

(divide by 4)

3

\int - sin x d x = - (- cos x) = cos x

4

\int cosec^{2} x d x = - cot x

Answer: $4 ln ∣ x ∣ + \frac{1}{4} e^{4 x} + cos x - cot x + c$

Practice

Question 2: Find

\int (3 cos 2 x + e^{- x} - 2 sec x tan x) d x

1

\int 3 cos 2 x d x = 3 \times \frac{1}{2} sin 2 x = \frac{3}{2} sin 2 x

2

\int e^{- x} d x = \frac{1}{- 1} e^{- x} = - e^{- x}

3

\int - 2 sec x tan x d x = - 2 sec x

Answer: $\frac{3}{2} sin 2 x - e^{- x} - 2 sec x + c$

Practice

Question 3: Find

\int (\frac{1}{2 x} + e^{3 x} - 4 sin 2 x + sec^{2} 3 x + cosec 2 x cot 2 x) d x

1

\int \frac{1}{2 x} d x = \frac{1}{2} ln ∣ x ∣

(or

\frac{1}{2} ln ∣2 x ∣ + c

)

2

\int e^{3 x} d x = \frac{1}{3} e^{3 x}

3

\int - 4 sin 2 x d x = - 4 \times (- \frac{1}{2} cos 2 x) = 2 cos 2 x

4

\int sec^{2} 3 x d x = \frac{1}{3} tan 3 x

5

\int cosec 2 x cot 2 x d x = - \frac{1}{2} cosec 2 x

Answer: $\frac{1}{2} ln ∣ x ∣ + \frac{1}{3} e^{3 x} + 2 cos 2 x + \frac{1}{3} tan 3 x - \frac{1}{2} cosec 2 x + c$

Summary

Standard Results to Memorise

FunctionIntegralFunctionIntegral

\frac{1}{x}

ln ∣ x ∣ + c

sec^{2} (k x)

\frac{1}{k} tan (k x) + c

e^{k x}

\frac{1}{k} e^{k x} + c

cosec^{2} (k x)

- \frac{1}{k} cot (k x) + c

sin (k x)

- \frac{1}{k} cos (k x) + c

sec (k x) tan (k x)

\frac{1}{k} sec (k x) + c

cos (k x)

\frac{1}{k} sin (k x) + c

cosec (k x) cot (k x)

- \frac{1}{k} cosec (k x) + c

Exam Tips

Always divide by k for functions of (kx)

Use modulus signs for

ln ∣ x ∣

Watch negative signs: sin, cosec², and cosec cot all have negatives

Check your answer by differentiating