Author: Rishabh Kumar (IIT + ISI | Global Math Mentor)
Published: October 2025
Category: Calculus | Integration Techniques
🔹 What Is Integration by Substitution?
Integration by Substitution is the reverse process of the chain rule in differentiation.
It’s used when the integral contains a composite function — something inside something else — and a direct integration approach doesn’t work cleanly.
The idea is simple:
Change the variable to make the integral simpler.
🔹 The Formula
If u=g(x)u = g(x)u=g(x), then dudx=g′(x)⇒du=g′(x) dx\frac{du}{dx} = g'(x) \quad \Rightarrow \quad du = g'(x)\,dxdxdu=g′(x)⇒du=g′(x)dx
Substituting into the integral: ∫f(g(x))g′(x) dx=∫f(u) du\int f(g(x))g'(x)\,dx = \int f(u)\,du∫f(g(x))g′(x)dx=∫f(u)du
Then integrate with respect to uuu, and finally substitute back u=g(x)u = g(x)u=g(x).
🔹 Step-by-Step Example 1 — Basic Substitution
∫2xcos(x2) dx\int 2x \cos(x^2)\,dx∫2xcos(x2)dx
Let ( u = x^2 \Rightarrow du = 2x,dx
]
So: ∫2xcos(x2) dx=∫cos(u) du=sin(u)+C=sin(x2)+C\int 2x \cos(x^2)\,dx = \int \cos(u)\,du = \sin(u) + C = \sin(x^2) + C∫2xcos(x2)dx=∫cos(u)du=sin(u)+C=sin(x2)+C
✅ Result: sin(x2)+C\sin(x^2) + Csin(x2)+C
This works because 2x2x2x was the derivative of x2x^2×2 — the inner function.
🔹 Example 2 — Using a Constant Multiplier
∫(3×2+2x)ex3+x2 dx\int (3x^2 + 2x) e^{x^3 + x^2}\,dx∫(3×2+2x)ex3+x2dx
Let ( u = x^3 + x^2 \Rightarrow du = (3x^2 + 2x),dx
] ∫(3×2+2x)ex3+x2 dx=∫eu du=eu+C=ex3+x2+C\int (3x^2 + 2x) e^{x^3 + x^2}\,dx = \int e^u\,du = e^u + C = e^{x^3 + x^2} + C∫(3×2+2x)ex3+x2dx=∫eudu=eu+C=ex3+x2+C
✅ Result: ex3+x2+Ce^{x^3 + x^2} + Cex3+x2+C
🔹 Example 3 — Trigonometric Case
∫sin(3x) dx\int \sin(3x)\,dx∫sin(3x)dx
Let ( u = 3x \Rightarrow du = 3,dx \Rightarrow dx = \frac{du}{3}
] ∫sin(3x) dx=13∫sin(u) du=−13cos(u)+C=−13cos(3x)+C\int \sin(3x)\,dx = \frac{1}{3}\int \sin(u)\,du = -\frac{1}{3}\cos(u) + C = -\frac{1}{3}\cos(3x) + C∫sin(3x)dx=31∫sin(u)du=−31cos(u)+C=−31cos(3x)+C
✅ Result: −13cos(3x)+C-\frac{1}{3}\cos(3x) + C−31cos(3x)+C
🔹 Example 4 — Substitution with Limits (Definite Integral)
∫0π/42sin(2x) dx\int_0^{\pi/4} 2\sin(2x)\,dx∫0π/42sin(2x)dx
Let ( u = 2x \Rightarrow du = 2dx \Rightarrow dx = \frac{du}{2}
]
Change limits:
- When x=0, u=0x = 0, \ u = 0x=0, u=0
- When x=π/4, u=π/2x = \pi/4, \ u = \pi/2x=π/4, u=π/2
Now: ∫0π/42sin(2x) dx=∫0π/2sin(u) du=[−cos(u)]0π/2=1\int_0^{\pi/4} 2\sin(2x)\,dx = \int_0^{\pi/2} \sin(u)\,du = [-\cos(u)]_0^{\pi/2} = 1∫0π/42sin(2x)dx=∫0π/2sin(u)du=[−cos(u)]0π/2=1
✅ Result: 111
🔹 Common Pitfalls
- ❌ Forgetting to substitute back u=g(x)u = g(x)u=g(x) at the end (for indefinite integrals).
- ❌ Forgetting to change the limits for definite integrals.
- ❌ Choosing the wrong substitution (make sure it simplifies the integral).
- ❌ Losing constants during substitution (dx=du/kdx = du/kdx=du/k).
🔹 Advanced Example — Reciprocal Function
∫x(x2+1)2 dx\int \frac{x}{(x^2 + 1)^2}\,dx∫(x2+1)2xdx
Let ( u = x^2 + 1 \Rightarrow du = 2x,dx \Rightarrow \frac{du}{2} = x,dx
] ∫x(x2+1)2 dx=12∫u−2 du=12(u−1−1)+C=−12u+C\int \frac{x}{(x^2 + 1)^2}\,dx = \frac{1}{2}\int u^{-2}\,du = \frac{1}{2}\left(\frac{u^{-1}}{-1}\right) + C = -\frac{1}{2u} + C∫(x2+1)2xdx=21∫u−2du=21(−1u−1)+C=−2u1+C
Substitute back: −12(x2+1)+C-\frac{1}{2(x^2 + 1)} + C−2(x2+1)1+C
✅ Result: −12(x2+1)+C-\frac{1}{2(x^2 + 1)} + C−2(x2+1)1+C
🔹 Connection with the Chain Rule
If: ddx[f(g(x))]=f′(g(x))g′(x)\frac{d}{dx}[f(g(x))] = f'(g(x))g'(x)dxd[f(g(x))]=f′(g(x))g′(x)
then ∫f′(g(x))g′(x) dx=f(g(x))+C\int f'(g(x))g'(x)\,dx = f(g(x)) + C∫f′(g(x))g′(x)dx=f(g(x))+C
Integration by substitution undoes the chain rule — reversing differentiation logic.
🔹 Real-World Applications
- Physics: velocity/acceleration under variable substitution
- Probability: transforming PDFs (change of variable)
- Economics: marginal functions and elasticity
- Advanced exams: IB HL, AP BC, A Level P3, STEP integrals
🌟 Why It Matters
Substitution transforms calculus from mechanical problem-solving into pattern recognition.
You start seeing structure — not just symbols.
It trains students to think like mathematicians: “What change of variable simplifies this?” — a mindset essential for higher-level exams and research.
📘 Learn Beyond the Formula
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In the Mathematics Elevate Mentorship Program, you’ll:
✅ Master substitution for all function types,
✅ Connect it with chain rule and geometry,
✅ Solve IB, AP, and STEP-level problems with confidence.
🚀 Transform how you approach integration.
👉 Book your personalized mentorship session now at MathByRishabh.com
