The chain rule differentiates composite functions: the derivative of sin(x2) is cos(x2)⋅2x — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given cos(x2)⋅2x, find the original function. That's what u substitution does — it reverses the chain rule.
The Core Intuition
When an integral looks like "a function times the derivative of its inside," substitute the inside with u and the derivative of the inside with du. Consider:
∫2xcos(x2)dx
Here 2x is the derivative of x2, and x2 is the inside of cos(x2). Let u=x2, so du=2xdx:
∫cos(u)du=sin(u)+C=sin(x2)+C
Check: the derivative of sin(x2) is cos(x2)⋅2x.
The Precise Statement
∫f(g(x))⋅g′(x)dx=∫f(u)duwhere u=g(x),du=g′(x)dx
Valid provided g is differentiable and the resulting integral in u is simpler.
The Step-by-Step Method
Identify a function g(x) whose derivative g′(x) also appears (possibly up to a constant factor).
Setu=g(x), compute du=g′(x)dx.
Rewrite the entire integral in u and du — every x and dx must be replaced.
Integrate with respect to u.
Substitute backu=g(x).
Watch out
You cannot mix variables. If any x remains after substitution, you chose the wrong u (or must solve for x in terms of u — rare).
A Second Example (with a constant factor)
Evaluate ∫xx2+1dx. Let u=x2+1, so xdx=21du:
∫u⋅21du=21⋅32u3/2+C=31(x2+1)3/2+C
When Does It Work?
When the integrand is something times the derivative of something inside. Common patterns:
x⋅f(x2) — derivative of x2 is 2x, so u=x2
eg(x)⋅g′(x) — derivative of g(x) appears
g(x)g′(x) — leads to log∣g(x)∣
Tip
If stuck, differentiate a candidate "inside" function in your head. If its derivative (up to a constant) appears, that's your u.
The Definite Integral Case
Either change the limits (when x=a, u=g(a); when x=b, u=g(b); then integrate in u), or integrate in u, substitute back, and use the original limits. Changing limits is cleaner:
Don't confuse du with Δu. du is a differential — the exact relationship du=g′(x)dx that holds inside the integral. Treat it algebraically: multiply, divide, and substitute freely.
U-substitution, taught in the CBSE Class 12 Integrals chapter as the method of substitution, is one of the very first integration techniques students learn after the standard formulas, and "integration by substitution class 12 examples" is a heavily searched revision topic. It remains equally essential for solving integral calculus problems in JEE Main and JEE Advanced.
The argument is linear (3x), so use the reverse chain rule (substitution) — no triple-angle identity is needed.
Recall ∫cos(ax)dx=a1sin(ax)+C. Here a=3:
∫cos3xdx=31sin3x+C.
Check: dxd(31sin3x)=31⋅3cos3x=cos3x.
✓Final answer
∫cos3xdx=31sin3x+C
Integrating cos3x is a linear-argument reverse chain rule: ∫cos3xdx=31sin3x+C.
The idea
We are integrating a cosine whose inside is 3x, not x. Guessing sin3x is close but wrong: differentiating sin3x gives 3cos3x — a factor of 3 too big. To undo that factor we divide by 3. (The triple-angle identity cos3x=4cos3x−3cosx is not the operative idea here and only complicates matters.)
Substitution
Let u=3x, so du=3dx, i.e. dx=3du. Then
∫cos3xdx=∫cosu⋅3du=31∫cosudu=31sinu+C.
Back-substitute
Replace u=3x:
∫cos3xdx=31sin3x+C.
Tip
General rule: ∫cos(ax+b)dx=a1sin(ax+b)+C — just divide by the coefficient of x.
✓Final answer
∫cos3xdx=31sin3x+C
Method: Integrating cos(ax+b) by the linear-argument rule
When a trig (or exponential) has a linear inside ax+b, no identity is needed — just divide by the coefficient of x.
Steps
Step 1: Recognise the linear argument.
cos3x has inner 3x; its antiderivative is a sine of the same argument, adjusted by the chain-rule factor.
Step 2: Apply the rule.
∫cos(ax+b)dx=a1sin(ax+b)+C.
Step 3: Verify.
Differentiate: dxd(a1sinax)=cosax. (Expanding cos3x=4cos3x−3cosx only complicates a one-step problem.)
Common Mistakes
Mistake 1: Forgetting the 31 factor.
Why it's wrong: dxdsin3x=3cos3x, so you must divide by 3. Correct approach: ∫cos3xdx=31sin3x+C, not sin3x+C.
Mistake 2: Expanding via the triple-angle identity.
Why it's wrong: it is unnecessary here — the argument is already linear. Correct approach: use the one-step rule ∫cos(ax)dx=a1sin(ax).