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 key idea is U Substitution — we substitute the inner expression to match the standard form ∫u2+11du=sinh−1u+C.
Let u=2−x. Then du=−dx, so dx=−du.
The integral becomes:
∫u2+11(−du)=−∫u2+11du
This integrates directly:
−sinh−1(u)+C
Substitute back u=2−x:
−sinh−1(2−x)+C
✓Final answer
The integral is −sinh−1(2−x)+C.
The key idea is to use a U-substitution that simplifies the denominator into a standard square-root-of-a-square form. By letting u=2−x, the integral becomes ∫u2+11du, which is a standard inverse hyperbolic sine (or sinh−1) result. The final answer is −sinh−1(2−x)+C.
Why U-Substitution Works Here
When you see an expression like (2−x)2+1, your first instinct should be: can I make this look like u2+1? That form is a classic — its integral is sinh−1u (or log∣u+u2+1∣, if you prefer). The inner function (2−x) is linear, so a simple substitution will cleanly transform the whole integrand.
The trap is to try expanding or completing the square — unnecessary. The structure is already perfect for a shift.
Step-by-Step Solution
Identify the substitution.
The troublesome part is (2−x). Let u=2−x. Then du=−dx, so dx=−du. This turns the integral into:
∫(2−x)2+11dx=∫u2+11(−du)=−∫u2+11du.
Recognize the standard form.
The integral ∫u2+11du is a known result. It equals sinh−1u+C (the inverse hyperbolic sine). If you haven't seen hyperbolic functions, the equivalent logarithmic form is:
sinh−1u=log(u+u2+1).
Either form is acceptable in Indian exams, but the sinh−1 notation is often preferred for brevity.
∫u2+a21du=sinh−1(au)+C(for a>0)
Here a=1, so it simplifies to sinh−1u+C.
Apply the result.
So:
−∫u2+11du=−sinh−1u+C.
Substitute back.
Recall u=2−x. Therefore:
∫(2−x)2+11dx=−sinh−1(2−x)+C.
Tip
If you prefer the logarithmic form, write:
−log(2−x+(2−x)2+1)+C.
Both are equivalent — use whichever your exam expects.
Watch out
A common mistake is forgetting the negative sign from dx=−du. Always check: if u=2−x, then du=−dx, so dx=−du. That minus sign must carry through to the final answer.
✓Final answer
The integral evaluates to −sinh−1(2−x)+C (or equivalently −log(2−x+(2−x)2+1)+C).
Method: Linear substitution into ∫u2+a2dx
A radical of the form (linear)2+a2 is reduced by substituting the linear part, turning it into a pure standard form.
Steps
Step 1: Substitute the linear inner expression.
For (2−x)2+1, set
u=2−x,du=−dx
Step 2: Rewrite the integral.
∫(2−x)2+1dx=−∫u2+1du
Step 3: Apply the standard result and restore.
Using ∫u2+a2du=logu+u2+a2+C (equivalently sinh−1au) with a=1, then substitute u=2−x back, carrying the minus sign from du=−dx.
The pattern: any (linear)2±a2 reduces to a standard form by substituting the linear block — just track the sign the differential introduces.
Common Mistakes
Mistake 1: Dropping the minus sign from du=−dx.
Why it's wrong: with u=2−x, du=−dx, so the integral picks up an overall minus; ignoring it flips the sign of the answer. Correct approach: ∫(2−x)2+1dx=−∫u2+1du=−sinh−1(2−x)+C.
Mistake 2: Not substituting the linear block and trying to expand (2−x)2.
Why it's wrong: expanding to x2−4x+5 obscures the ready-made u2+1 form. Correct approach: substitute u=2−x directly since the square is already completed.