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:
The integral ∫0π/2sin3xdx is evaluated by rewriting sin3x as sinx(1−cos2x) and using the substitution u=cosx, which transforms the integral into a simple polynomial form. The value is 32.
The key to evaluating powers of sine or cosine over a symmetric interval like [0,π/2] is often to use a trigonometric identity to reduce the power, then substitute. For sin3x, the direct approach is to factor it as sinx⋅sin2x, then replace sin2x with 1−cos2x. This sets up a perfect substitution because the derivative of cosx is −sinx, which appears as a factor.
Let’s work through it step by step.
Rewrite the integrand
We have sin3x=sinx⋅sin2x=sinx(1−cos2x).
So the integral becomes
I=∫0π/2sinx(1−cos2x)dx.
Choose a substitution
Let u=cosx. Then du=−sinxdx, so sinxdx=−du.
When x=0, u=cos0=1. When x=π/2, u=cos(π/2)=0.
The limits reverse: the lower limit becomes u=1 and the upper limit becomes u=0.
Why it's wrong: sin3x has no direct antiderivative until you peel one sinx and use sin2x=1−cos2x. Correct approach: write sin3x=sinx(1−cos2x).
Mistake 2: Sign/limit slip from u=cosx.
Why it's wrong: du=−sinxdx, and the limits reverse (x=0→u=1, x=2π→u=0); mishandling either gives the wrong sign. Correct approach: use sinxdx=−du and flip the limits. …
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
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CBSE 2026Set 65/1/11 markMCQ
Q.If ∫b2+c2x23axdx=Alog∣b2+c2x2∣+K, then the value of A is:
(A) 3a
(B) 2b23a
(C) b2c23a
(D) 2c23a
›Reveal solutionSolution
The integral fits the pattern ∫udu=log∣u∣+C after a substitution. The constant A turns out to be 2c23a, which corresponds to option (D).
The problem gives you the result of an integral and asks you to identify the constant A that makes the equation true. This is a classic "match the form" question — you don't need to guess; you just need to perform the integration carefully and compare.
The key insight is that the integrand b2+c2x23ax is a rational function where the numerator is almost the derivative of the denominator. The derivative of b2+c2x2 is 2c2x. Our numerator is 3ax, which is a constant multiple of x. So a simple substitution u=b2+c2x2 will turn the integral into ∫udu.
Let's work through it step by step.
Set up the substitution.
Let u=b2+c2x2. Then du=2c2xdx, so xdx=2c2du.
Rewrite the integral in terms of u.
The integral is ∫b2+c2x23axdx=∫u3a⋅(xdx).
Substitute xdx=2c2du:
∫u3a⋅2c2du=2c23a∫udu.
Integrate.∫udu=log∣u∣+C, so
2c23alog∣u∣+C=2c23alog∣b2+c2x2∣+K,
where K is the constant of integration (we renamed C to K to match the problem).