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Exercise 7.1 · Q2

Q.Integrate the following function: cos⁡3x\cos 3x

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Concept understanding — U Substitution

U Substitution: The Reverse Chain Rule

The chain rule differentiates composite functions: the derivative of sin⁡(x2)\sin(x^2) is cos⁡(x2)⋅2x\cos(x^2) \cdot 2x — differentiate the outer function, then multiply by the derivative of the inside. Integration asks the reverse: given cos⁡(x2)⋅2x\cos(x^2) \cdot 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 uu and the derivative of the inside with dudu. Consider:

∫2xcos⁡(x2) dx\int 2x \cos(x^2) \, dx

Here 2x2x is the derivative of x2x^2, and x2x^2 is the inside of cos⁡(x2)\cos(x^2). Let u=x2u = x^2, so du=2x dxdu = 2x \, dx:

∫cos⁡(u) du=sin⁡(u)+C=sin⁡(x2)+C\int \cos(u) \, du = \sin(u) + C = \sin(x^2) + C

Check: the derivative of sin⁡(x2)\sin(x^2) is cos⁡(x2)⋅2x\cos(x^2) \cdot 2x.

The Precise Statement

∫f(g(x))⋅g′(x) dx=∫f(u) duwhere u=g(x), du=g′(x) dx\int f(g(x)) \cdot g'(x) \, dx = \int f(u) \, du \quad \text{where } u = g(x), \, du = g'(x) \, dx

Valid provided gg is differentiable and the resulting integral in uu is simpler.

The Step-by-Step Method

  1. Identify a function g(x)g(x) whose derivative g′(x)g'(x) also appears (possibly up to a constant factor).
  2. Set u=g(x)u = g(x), compute du=g′(x) dxdu = g'(x) \, dx.
  3. Rewrite the entire integral in uu and dudu — every xx and dxdx must be replaced.
  4. Integrate with respect to uu.
  5. Substitute back u=g(x)u = g(x).
Watch out

You cannot mix variables. If any xx remains after substitution, you chose the wrong uu (or must solve for xx in terms of uu — rare).

A Second Example (with a constant factor)

Evaluate ∫xx2+1 dx\int x \sqrt{x^2 + 1} \, dx. Let u=x2+1u = x^2 + 1, so x dx=12dux \, dx = \frac{1}{2} du:

∫u⋅12du=12⋅23u3/2+C=13(x2+1)3/2+C\int \sqrt{u} \cdot \tfrac{1}{2} du = \tfrac{1}{2} \cdot \tfrac{2}{3} u^{3/2} + C = \tfrac{1}{3} (x^2 + 1)^{3/2} + C

When Does It Work?

When the integrand is something times the derivative of something inside. Common patterns:

  • x⋅f(x2)x \cdot f(x^2) — derivative of x2x^2 is 2x2x, so u=x2u = x^2
  • eg(x)⋅g′(x)e^{g(x)} \cdot g'(x) — derivative of g(x)g(x) appears
  • g′(x)g(x)\frac{g'(x)}{g(x)} — leads to log⁡∣g(x)∣\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 uu.

The Definite Integral Case

Either change the limits (when x=ax = a, u=g(a)u = g(a); when x=bx = b, u=g(b)u = g(b); then integrate in uu), or integrate in uu, substitute back, and use the original limits. Changing limits is cleaner:

∫x=0x=12xcos⁡(x2) dx=∫u=0u=1cos⁡(u) du=sin⁡(1)−sin⁡(0)=sin⁡(1)\int_{x=0}^{x=1} 2x \cos(x^2) \, dx = \int_{u=0}^{u=1} \cos(u) \, du = \sin(1) - \sin(0) = \sin(1)

Common Mistake to Avoid

Don't confuse dudu with Δu\Delta u. dudu is a differential — the exact relationship du=g′(x)dxdu = 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.

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