Substitution method
To convert a long bill into a simple one you may count in bundles instead of single items. Substitution bundles a complicated part of the integrand into a single new letter.
If the integrand has the form f(g(x)) · g′(x), put t = g(x). Then g′(x) dx becomes dt, and the integral becomes ∫ f(t) dt, which is easy. At the end, write t back in terms of x.
Formula: ∫ f(g(x)) g′(x) dx = ∫ f(t) dt, where t = g(x)
Rule 1: Look for a function and its derivative — Choose t as the inside part.
- Easy: ∫ 2x (x² + 1)³ dx: put t = x² + 1, dt = 2x dx, giving t⁴/4 + C = (x² + 1)⁴/4 + C.
- Medium: ∫ cos 3x dx: put t = 3x, dx = dt/3, giving (sin 3x)/3 + C.
- Exam-type: ∫ x e^(x²) dx: put t = x², x dx = dt/2, giving e^(x²)/2 + C.
Rule 2: Linear inside — For f(ax + b) divide by a.
- Easy: ∫ (2x + 3)⁴ dx = (2x + 3)⁵ / 10 + C.
- Medium: ∫ e⁵ˣ dx = e⁵ˣ / 5 + C.
- Exam-type: ∫ x √(x² + 1) dx = (x² + 1)^(3/2) / 3 + C.
After substituting, the integral must contain only t. If an x is left over, the choice of t was not right.
Forgetting to convert dx into dt, or leaving the answer in t.
Inside part becomes t, its derivative becomes dt.
Practice questions
- 1Find ∫ (2x + 3)⁴ dx.
- 2Find ∫ x (x² + 4)⁵ dx.
- 3Find ∫ e⁵ˣ dx.
- 4Find ∫ sin 2x dx.
- 5Find ∫ 3x² / (x³ + 1) dx.
- 6Find ∫ x √(x² + 1) dx.
- 7Find ∫ (ln x)² / x dx.
- 8Find ∫ x² e^(x³) dx.
Show answers (try them yourself first)
- 1(2x + 3)⁵ / 10 + C
- 2(x² + 4)⁶ / 12 + C
- 3e⁵ˣ / 5 + C
- 4−cos 2x / 2 + C
- 5ln|x³ + 1| + C
- 6(x² + 1)^(3/2) / 3 + C
- 7(ln x)³ / 3 + C
- 8e^(x³) / 3 + C
Standard results found by substitution
A small trick saves a lot of time. Four integrals of trigonometric functions come up again and again, so they are worth keeping in the memory.
Writing tan x = sin x / cos x and putting t = cos x gives the integral of tan x. The same idea gives the other three results.
Formula: ∫ tan x dx = ln|sec x| + C, ∫ cot x dx = ln|sin x| + C, ∫ sec x dx = ln|sec x + tan x| + C, ∫ cosec x dx = ln|cosec x − cot x| + C
Rule 1: Numerator is the derivative of denominator — The answer is ln of the denominator.
- Easy: ∫ cot x dx = ln|sin x| + C.
- Medium: ∫ cos x / (1 + sin x) dx = ln|1 + sin x| + C.
- Exam-type: ∫ eˣ / (1 + eˣ) dx = ln(1 + eˣ) + C.
Rule 2: Scaled angles — For tan 3x, sec 2x and so on, divide by the coefficient of x.
- Easy: ∫ tan 3x dx = (1/3) ln|sec 3x| + C.
- Medium: ∫ sec x dx = ln|sec x + tan x| + C.
- Exam-type: ∫ cosec 2x dx = (1/2) ln|cosec 2x − cot 2x| + C.
When the numerator is the derivative of the denominator, the integral is ln|denominator| + C.
Writing ∫ tan x dx = sec²x or (tan²x)/2.
Top is the slope of the bottom, so take the log.
Practice questions
- 1Find ∫ cot x dx.
- 2Find ∫ tan 3x dx.
- 3Find ∫ sec x dx.
- 4Find ∫ cosec 2x dx.
- 5Find ∫ sec²x / tan x dx.
- 6Find ∫ cos x / (1 + sin x) dx.
- 7Find ∫ eˣ / (1 + eˣ) dx.
- 8Find ∫ sin x cos x dx using t = sin x.
Show answers (try them yourself first)
- 1ln|sin x| + C
- 2(1/3) ln|sec 3x| + C
- 3ln|sec x + tan x| + C
- 4(1/2) ln|cosec 2x − cot 2x| + C
- 5ln|tan x| + C
- 6ln|1 + sin x| + C
- 7ln(1 + eˣ) + C
- 8sin²x / 2 + C