Integration Substitution Techniques
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Direct answer
Substitution runs the chain rule backwards: in ∫ f(g(x))·g'(x) dx, set t = g(x) so the integral collapses to ∫f(t) dt, and the whole craft is spotting which composite to kill. Quadratic radicals dictate trigonometric substitutions — x = a sin θ for √(a² − x²), x = a tan θ for √(a² + x²), x = a sec θ for √(x² − a²) — because each choice turns the surd into a single trigonometric factor. Rational functions of sin x and cos x yield to the Weierstrass half-angle t = tan(x/2), with sin x = 2t/(1 + t²), cos x = (1 − t²)/(1 + t²), dx = 2 dt/(1 + t²), which rationalises every such integrand. Definite integrals demand the extra discipline of transforming the limits along with the variable.
What you must remember
- Linear composites: ∫f(ax + b)dx = F(ax + b)/a — the most automatic substitution; always divide by the inner derivative.
- Trigonometric substitution table: √(a² − x²) → x = a sin θ (cos θ survives); √(a² + x²) → x = a tan θ (sec θ survives); √(x² − a²) → x = a sec θ (tan θ survives).
- Back-substitution discipline: after integrating in θ, express the result in x using a reference triangle; a final answer still in θ loses the mark in school-style marking.
- Half-angle substitution: t = tan(x/2) rationalises any integrand rational in sin x and cos x; special faster forms exist when the integrand is odd in sin (use t = cos x) or odd in cos (use t = sin x).
- Reciprocal trick: for symmetric forms like (1 + x²)/(1 + x⁴), t = x − 1/x or t = x + 1/x exploits the balanced exponents.
- Definite-integral rule: change the limits when you change the variable — a transformed integrand with untransformed limits is the single most expensive careless error.
- Standard results unlocked: ∫dx/(a² + x²) = (1/a)tan⁻¹(x/a) + C and ∫dx/√(a² − x²) = sin⁻¹(x/a) + C are themselves substitution results.
Half-angle substitution, end to end
Evaluate ∫dx/(1 + cos x). Set t = tan(x/2), so cos x = (1 − t²)/(1 + t²) and dx = 2dt/(1 + t²). Then 1 + cos x = 2/(1 + t²), and the integral becomes ∫[2/(1 + t²)]·[(1 + t²)/2]dt = ∫dt = t + C = tan(x/2) + C. One substitution removed both the cosine and the differential at once.
The trigonometric-substitution branch shows its own character on ∫dx/((x² + 4)^(3/2)): set x = 2 tan θ, so x² + 4 = 4 sec²θ and dx = 2 sec²θ dθ. The integrand becomes 2 sec²θ/(8 sec³θ) dθ = (1/4)cos θ dθ = (1/4)sin θ + C = x/(4√(x² + 4)) + C after back-substitution through the reference triangle (opposite x, adjacent 2, hypotenuse √(x² + 4)) — the surd dissolved into a secant identity, the entire purpose of the substitution table.
Choosing the substitution
Diagnostic questions precede technique. Is the integrand a function of a linear inner term? Divide by the derivative, done. Is there a quadratic surd? Match a² − x², a² + x², x² − a² to its substitution. Is it rational in sin and cos? Test parity first (odd in sin → t = cos x is faster), and half-angle only otherwise. Are exponents symmetric in x and 1/x? The reciprocal trick applies. Main tests recognition of the standard three; Advanced composes — surds inside rational functions, or symmetry properties like ∫₀^(π/2) f(sin x)dx = ∫₀^(π/2) f(cos x)dx, which are substitution arguments in disguise. The standing traps: untransformed definite-integral limits, a missing +C, and the half-angle blind spot at x = π, where tan(x/2) is undefined — split any interval crossing π.
Frequently asked questions
Which trigonometric substitution fits √(a² + x²)?
x = a tan θ, since a² + a² tan²θ = a² sec²θ and the square root collapses to a sec θ.
What is the Weierstrass substitution and when is it used?
t = tan(x/2), giving sin x = 2t/(1 + t²), cos x = (1 − t²)/(1 + t²), dx = 2dt/(1 + t²); it converts any rational function of sin x and cos x into a rational function of t.
What must accompany a substitution in a definite integral?
The limits must be transformed with the variable; if t = g(x), then x = a → t = g(a) and x = b → t = g(b).
When can t = sin x or t = cos x replace the half-angle substitution?
When the integrand is odd in cos x (t = sin x works) or odd in sin x (t = cos x works) — factoring out one power of the odd function leaves an even remainder in the other.
How do you convert the θ-answer back to x after trigonometric substitution?
Build the reference triangle from the substitution (for x = a tan θ: opposite x, adjacent a), read sin θ, cos θ or their reciprocals as side ratios, and substitute — never leave θ in the final answer.