Standard Forms of Trigonometric Integration
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Direct answer
Every rational function of sin x and cos x yields to the single substitution t = tan(x/2): sin x = 2t/(1 + t^2), cos x = (1 - t^2)/(1 + t^2), dx = 2 dt/(1 + t^2), which converts the integrand into a rational function of t. When the integrand is linear in sine and cosine, as in dx/(a sin x + b cos x), the faster route writes a sin x + b cos x = R sin(x + φ) with R = √(a^2 + b^2) and tan φ = b/a. Power integrals sin^m x cos^n x split by parity: an odd power of either factor invites a substitution (u = sin x or u = cos x) after peeling off one factor, while two even powers demand the half-angle demotions sin^2 x = (1 - cos 2x)/2 and cos^2 x = (1 + cos 2x)/2.
What you must remember
- Weierstrass substitution: t = tan(x/2) with sin x = 2t/(1 + t^2), cos x = (1 - t^2)/(1 + t^2), dx = 2 dt/(1 + t^2); use it as the universal fallback when no structure is visible.
- Key result: ∫dx/(a + b cos x) for a > b > 0 equals 2/√(a^2 - b^2) × arctan(√((a - b)/(a + b)) tan(x/2)) + C; for a < b the answer turns logarithmic.
- Phase-shift trick: a sin x + b cos x = R sin(x + φ) with R = √(a^2 + b^2), tan φ = b/a; then ∫dx/(a sin x + b cos x) = (1/R) ln|tan((x + φ)/2)| + C.
- Odd-power strategy: for sin^m x cos^n x with m odd, keep one sin x aside and convert the rest to cosines, substituting u = cos x; mirror for n odd.
- Even-power strategy: both even means repeated half-angle demotion to cos 2x and cos 4x.
- Standard answers worth memorising: ∫dx/(1 + sin x) = tan x - sec x + C; ∫dx/(sin x + cos x) = (1/√2) ln|tan(x/2 + π/8)| + C.
- Radical forms: ∫√((1 - cos x)/(1 + cos x)) dx = ∫tan(x/2) dx = 2 ln|sec(x/2)| + C, a classic NCERT-derived ask.
One integral, two routes
Compute ∫dx/(1 + sin x). Route one, the conjugate: multiply above and below by (1 - sin x), so the integrand becomes (1 - sin x)/cos^2 x = sec^2 x - sec x tan x. Each piece is a standard derivative backwards: sec^2 x integrates to tan x and sec x tan x to sec x, giving tan x - sec x + C. Route two, Weierstrass: with t = tan(x/2) the integrand becomes [2/(1 + t^2)] ÷ [1 + 2t/(1 + t^2)] = 2/(t^2 + 2t + 1) = 2/(t + 1)^2, which integrates to -2/(t + 1) + C. Differentiate tan x - sec x (getting sec^2 x - sec x tan x, which rearranges to 1/(1 + sin x)) or substitute t = tan(x/2) into -2/(t + 1) — both routes land on equivalent answers differing by a constant. The lesson generalises: when the integrand is a simple combination of first powers, conjugates and phase shifts beat the heavy substitution; when powers pile up, t = tan(x/2) always works but always costs algebra.
Where students slip
JEE Main draws heavily from the standard-form shelf: ∫dx/(5 + 4 cos x) type results (answer 2/3 arctan((1/3)tan(x/2))), power integrals with one odd exponent, and the sec-tan family — usually as definite-integral numericals. Advanced mixes the substitution into larger structures: definite integrals over [0, π/2] where king-property symmetry makes t = tan(x/2) unnecessary, or integrals where the wrong choice of route buries you in partial fractions. The common failures: applying t = tan(x/2) to definite integrals and forgetting that the limits must also change (x: 0 → π means t: 0 → ∞); writing √(a^2 + b^2) as the amplitude but forgetting φ's quadrant (tan φ = b/a does not fix the sign of cos φ); and in odd-power problems, peeling a sin x but converting the remainder into cosines while still writing du = cos x dx on the wrong side. Trigonometric integration is core calculus syllabus for both exams and among the highest-frequency integral types in JEE Main.
Frequently asked questions
When should t = tan(x/2) be used?
As the universal substitution for any rational function of sin x and cos x — especially when both appear with higher powers or no conjugate trick is visible; remember to transform the limits in definite integrals.
How is a sin x + b cos x simplified?
Write it as R sin(x + φ) with R = √(a^2 + b^2) and tan φ = b/a, turning the integrand into a single shifted sine.
What is the strategy when both powers in sin^m x cos^n x are even?
Demote with half-angle formulas sin^2 x = (1 - cos 2x)/2 and cos^2 x = (1 + cos 2x)/2, repeatedly if necessary, until only cos 2x and cos 4x terms remain.
What is ∫dx/(1 + sin x)?
tan x - sec x + C, obtained by multiplying with the conjugate (1 - sin x); the Weierstrass route gives -2/(1 + tan(x/2)) + C, the same family.
Which form does ∫dx/(a + b cos x) take when a > b?
2/√(a^2 - b^2) × arctan(√((a - b)/(a + b)) tan(x/2)) + C; when b > a the arctangent becomes a logarithm.