Integration by substitution
Put t = g(x), so dt = g′(x) dx. The integral in x becomes a standard integral in t. Integrate, then put back x.
How to substitute
Use substitution when one part of the integrand is the derivative of another part, up to a constant factor.
- 1Spot a function g(x) whose derivative g′(x) also appears as a factor.
- 2Put t = g(x). Differentiate: dt = g′(x) dx.
- 3Replace every x and dx. No x should be left.
- 4Integrate in t using a standard integral.
- 5Put back t = g(x) and add C.
- 6Check: differentiate the answer with the chain rule. You should get the integrand.
Results proved by substitution
| Integral | Substitution | Result |
|---|---|---|
| ∫ tan x dx | t = cos x, dt = −sin x dx | −log|cos x| + C = log|sec x| + C |
| ∫ cot x dx | t = sin x, dt = cos x dx | log|sin x| + C |
| ∫ sec x dx | Multiply top and bottom by (sec x + tan x); t = sec x + tan x | log|sec x + tan x| + C |
| ∫ cosec x dx | Multiply top and bottom by (cosec x − cot x); t = cosec x − cot x | log|cosec x − cot x| + C |
| ∫ f′(x)/f(x) dx | t = f(x) | log|f(x)| + C |
| ∫ [f(x)]ⁿ f′(x) dx, n ≠ −1 | t = f(x) | [f(x)]ⁿ⁺¹/(n + 1) + C |
Use a trigonometric identity first
- Squares: sin² x = (1 − cos 2x)/2 and cos² x = (1 + cos 2x)/2. So ∫ cos² x dx = x/2 + (sin 2x)/4 + C.
- Cubes: sin³ x = (3 sin x − sin 3x)/4 and cos³ x = (3 cos x + cos 3x)/4.
- Products: 2 sin A cos B = sin(A + B) + sin(A − B); 2 cos A cos B = cos(A + B) + cos(A − B); 2 sin A sin B = cos(A − B) − cos(A + B).
- So ∫ sin 3x cos 2x dx = (1/2) ∫ (sin 5x + sin x) dx = −(cos 5x)/10 − (cos x)/2 + C.
- Odd power of sin x or cos x: keep one factor aside, change the rest with sin² x + cos² x = 1, then put t = cos x or t = sin x.
Example 1: spot the derivative
Find ∫ cos(√x)/√x dx.
- The derivative of √x is 1/(2√x), and 1/√x appears in the integrand.
- Put t = √x. Then dt = dx/(2√x), so dx/√x = 2 dt.
- The integral becomes 2 ∫ cos t dt.
- = 2 sin t + C.
- Put back t = √x.
- Check: d/dx [2 sin √x] = 2 cos √x × 1/(2√x) = cos √x/√x, the integrand.
2 sin √x + C
Example 2: identity, then substitution
Find ∫ sin³ x cos⁴ x dx.
- The power of sin x is odd, so keep one sin x aside: sin³ x = sin² x × sin x.
- Change sin² x = 1 − cos² x. The integrand is (1 − cos² x) cos⁴ x sin x.
- Put t = cos x. Then dt = −sin x dx, so sin x dx = −dt.
- The integral becomes −∫ (1 − t²) t⁴ dt = −∫ (t⁴ − t⁶) dt.
- = −t⁵/5 + t⁷/7 + C.
- Put back t = cos x.
−(cos⁵ x)/5 + (cos⁷ x)/7 + C
Changing the variable is like converting a recipe from cups to grams: you work in the easier unit, then convert back. Engineers do the same when they switch from time to the angle turned by a rotating fan or wheel.
2 marks: “Find ∫ (2 + log x)³/x dx.” Put t = 2 + log x, dt = dx/x. ∫ t³ dt = t⁴/4 + C. Answer: (2 + log x)⁴/4 + C.
2 marks: “Find ∫ dx/(x(2 + 3 log x)).” Put t = 2 + 3 log x, dt = (3/x) dx. (1/3) ∫ dt/t = (1/3) log|t| + C. Answer: (1/3) log|2 + 3 log x| + C.
3 marks: “Find ∫ cos x/cos(x − a) dx.” Put x − a = t, dx = dt. cos x = cos(t + a) = cos t cos a − sin t sin a, so the integrand = cos a − sin a tan t. Integral = t cos a + sin a log|cos t| + C₁. Put back t = x − a and absorb −a cos a into the constant: x cos a + sin a log|cos(x − a)| + C.
Spot the derivative. If the derivative of one part sits next to it as a factor, that part is your t.