Integration by Substitution: JEE Main Guide 2026

Integration by substitution is the most powerful single technique in the JEE Main integration toolkit. Approximately 40–50% of all indefinite integration questions in JEE can be solved by substitution alone — the others require by-parts, partial fractions, or a combination. Mastering substitution — knowing which substitution to make for which integrand family — is the single highest-return investment in Calculus preparation. This guide systematises every major substitution type the exam uses.

Direct Substitution: The Simplest Family

Direct substitution: when the integrand contains f(g(x))·g'(x), substitute u=g(x): ∫f(g(x))g'(x)dx = ∫f(u)du. Recognition: look for a composite function where the derivative of the inner function also appears. Examples: ∫2x(x²+1)⁵dx — substitute u=x²+1, du=2x dx: ∫u⁵du = u⁶/6. ∫cos(3x+2)dx — substitute u=3x+2, du=3dx: (1/3)∫cosu du = (1/3)sinu = (1/3)sin(3x+2). ∫e^(sinx)·cosx dx — substitute u=sinx: ∫eᵘdu = eᵘ = e^(sinx). The reflex to build: see a composite function, look for its derivative, make the substitution.

Slightly less direct: ∫f(ax+b)dx = (1/a)F(ax+b)+C where F is the antiderivative of f. This linear function rule covers: ∫sin(5x−3)dx = −(1/5)cos(5x−3), ∫e^(2x+1)dx = (1/2)e^(2x+1), ∫(3x+1)⁴dx = (1/15)(3x+1)⁵. JEE uses this in multi-step integrals where the first step is this linear rule. Test your substitution fluency with a free integration mock. For the full integration framework, see our integration techniques guide.

Trigonometric Substitutions for Irrational Integrands

Standard substitutions for integrands containing square roots: √(a²−x²) → substitute x = a sinθ (transforms to √(a²cos²θ) = a cosθ). √(a²+x²) → substitute x = a tanθ (transforms to a secθ). √(x²−a²) → substitute x = a secθ (transforms to a tanθ). After integrating in terms of θ, back-substitute: sinθ = x/a → θ = sin⁻¹(x/a), etc. These three standard forms cover all irrational algebraic integrands of this type.

For integrands of the form 1/(ax²+bx+c): complete the square to get 1/((x+p)²±q²), then apply the standard result ∫dx/(x²+a²) = (1/a)tan⁻¹(x/a)+C or ∫dx/(x²−a²) = (1/2a)ln|(x−a)/(x+a)|+C. For √(ax²+bx+c): complete the square, factor out √a, then use trig substitution. These algebraic rearrangement steps — completing the square — precede the trig substitution and are where most errors occur. For related definite integration techniques, see our definite integrals guide.

Weierstrass and Euler Substitutions for Trig Integrands

For ∫R(sinx, cosx)dx where R is a rational function of sinx and cosx: use the Weierstrass substitution t = tan(x/2), giving sinx = 2t/(1+t²), cosx = (1−t²)/(1+t²), dx = 2dt/(1+t²). This converts the integral into a rational function of t, solvable by partial fractions. Applications: ∫dx/(2+sinx), ∫dx/(3+5cosx), ∫dx/(a+b sinx) — all standard JEE types. The pattern-recognition shorthand: if denominator is a linear function of sinx and/or cosx, use Weierstrass.

Special cases that do not need Weierstrass: ∫sinᵐx cosⁿx dx. If m or n is odd (say m is odd): substitute u=cosx. If both are even: use double-angle formulas first (sin²x = (1−cos2x)/2). If m=n=even: this is the most work-intensive case and uses multiple double-angle reductions. JEE avoids the hardest cases (both even and large) but does use m odd or n odd with regularity. For integration by parts connections, see our integration by parts guide.

Partial Fraction Substitution for Rational Functions

For ∫P(x)/Q(x)dx where deg(P) < deg(Q): factorise Q(x), then decompose into partial fractions. For irreducible linear factors (x−a): A/(x−a). For repeated linear (x−a)²: A/(x−a)+B/(x−a)². For irreducible quadratic (x²+bx+c): (Cx+D)/(x²+bx+c). JEE most commonly uses denominator forms: (x−a)(x−b), (x−a)(x²+1), (x−a)²(x−b). After decomposition, integrate each fraction using standard results. The "cover-up" method for linear factors: to find A in A/(x−a), cover (x−a) in Q(x) and substitute x=a into P(x)/[remaining Q(x)]. This method is twice as fast as comparing coefficients for simple poles. For the full partial fractions coverage, see our integration partial fractions guide.