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 ∫1−x8x3dx is solved by the substitution u=x4, which transforms it into a standard arcsine form. The final result is 41sin−1(x4)+C.
The key insight here is that the denominator contains 1−x8, and x8=(x4)2. That square inside a square root under 1−(something)2 is a dead giveaway for the arcsine derivative formula: dudsin−1u=1−u21.
But we have x3 in the numerator, not x4 or something that directly matches. That’s where substitution comes in — we need to turn the numerator into the derivative of the “something” we want to put inside the arcsine.
Choose the substitution.
Let u=x4. Then du=4x3dx, so x3dx=4du.
Why x4? Because x8=(x4)2=u2, and the numerator x3 is exactly the derivative of x4 up to a constant factor. This is the cleanest way to match the arcsine form.
Rewrite the integral.
The original integral is
∫1−x8x3dx=∫1−(x4)21⋅x3dx.
Substituting u=x4 and x3dx=4du gives
∫1−u21⋅4du=41∫1−u2du.
Recognise the standard integral.
The integral ∫1−u2du is exactly sin−1u+C. This is a fundamental result from differentiation: dudsin−1u=1−u21.
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).