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 ∫xcosxdx is solved by substituting u=x, which simplifies the integrand to 2cosu. The final answer is 2sinx+C.
The key to this problem is noticing that the derivative of x is 2x1, and we have a x1 factor sitting right next to cosx. That’s a dead giveaway for U Substitution — we let the inner function be u, and the rest of the integrand becomes its derivative (up to a constant factor).
Let’s walk through it.
Set up the substitution.
Let u=x. Then x=u2, so dx=2udu.
Why this choice? Because x appears inside the cosine, and its derivative 2x1 is almost exactly the x1 we have. The substitution will collapse the whole expression into something clean.
Rewrite the integral in terms of u.
The original integral is
∫xcosxdx.
Replace x with u, and dx with 2udu:
∫xcosu⋅2udu.
But x=u, so x1=u1. That gives:
∫ucosu⋅2udu=∫2cosudu.
The u cancels beautifully — that’s the whole point of the substitution.
Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.
Showing the 12 most recent of 44 on this concept.
CBSE 2020Set 65/1/11 mark
Q.Find : ∫9−4x2dx
›Reveal solutionSolution
The integral ∫9−4x2dx is a standard inverse sine form. By rewriting the denominator as 4(49−x2) and using substitution u=2x, we get the result 21sin−1(32x)+C.
When you see a square root with a constant minus a square term, your mind should immediately jump to the inverse trigonometric integrals. The classic formula is:
∫a2−u2du=sin−1(au)+C
Our job is to force the given integral into this exact shape. The denominator is 9−4x2. Notice that 9=32, so we have a=3 in the formula. But the 4x2 term is not a pure u2 — it has a coefficient 4. That’s the only obstacle.
The key insight: Factor out the 4 from inside the square root. Write:
9−4x2=4(49−x2)=249−x2
Now the integral becomes:
∫249−x2dx=21∫(23)2−x2dx
This is exactly the inverse sine form with a=23 and u=x. So:
21sin−1(3/2x)+C=21sin−1(32x)+C
That’s the answer. But let’s walk through it step by step with a substitution to make it foolproof.
Identify the target form. We want ∫a2−u2du. Here, the denominator has 9−4x2. Compare with a2−u2: we need a2=9 and u2=4x2. So set u=2x. Then du=2dx, so dx=2du.
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).