Using substitution to integrate composite functions, reversing the chain rule.
Answer at least 3 of 3 correctly to complete this section.
You already know how to integrate polynomials using the reverse power rule. But what about ∫x(x2+1)4dx or ∫sin2xcosxdx? These do not fit the basic rules directly.
Integration by substitution is the reverse of the chain rule. Just as the chain rule lets you differentiate composite functions, substitution lets you integrate them. It works by replacing a complicated expression with a single variable u, turning a hard integral into an easy one.
This technique appears on almost every A-Level Paper 1. Sometimes the substitution is given; sometimes you must choose it yourself. Either way, the method is the same — and once you see the pattern, it becomes second nature.
When you see a composite function inside an integral, you can simplify it by letting u equal the inner function. The key steps are:
Find ∫(2x+3)5dx.
Let u=2x+3. Then dxdu=2, so dx=2du.
∫(2x+3)5dx=∫u5⋅2du=21∫u5du
=21⋅6u6+c=12u6+c=12(2x+3)6+c
Check by differentiating: dxd[12(2x+3)6]=126(2x+3)5⋅2=(2x+3)5 ✓
Find ∫2x(x2+1)4dx.
The inner function is x2+1, and its derivative 2x appears in the integrand. This is the perfect signal for substitution.
Let u=x2+1. Then dxdu=2x, so du=2xdx.
∫2x(x2+1)4dx=∫u4du=5u5+c=5(x2+1)5+c
Notice how neatly the 2xdx became du. When the derivative of the inner function appears as a factor in the integrand, the substitution works perfectly.
Find ∫x(x2+1)4dx.
This time we have x rather than 2x. Let u=x2+1, so du=2xdx.
We need xdx, which is 21du:
∫x(x2+1)4dx=∫u4⋅2du=21⋅5u5+c=10(x2+1)5+c
Key point: You can adjust for constant multiples (multiply/divide by a number), but you cannot adjust for functions of x. If there is an x left over that does not cancel, you need a different approach.
Answer at least 3 of 3 correctly to complete this section.
For definite integrals, you have two options:
Option 1 is cleaner and less error-prone.
Evaluate ∫02xx2+1dx.
Let u=x2+1, so du=2xdx, giving xdx=2du.
Change the limits:
∫02xx2+1dx=∫15u⋅2du=21∫15u1/2du
=21[3/2u3/2]15=21⋅32[u3/2]15=31(53/2−1)
=31(55−1)
Common error: Students change the integrand to u but leave the limits as x=0 and x=2. If you work in u, the limits must be u-values. If you want to keep the x-limits, you must substitute back to x before evaluating.
Using the substitution u=sinx, find ∫0π/2sin3xcosxdx.
With u=sinx: dxdu=cosx, so du=cosxdx.
Change the limits:
∫0π/2sin3xcosxdx=∫01u3du=[4u4]01=41
Answer at least 3 of 3 correctly to complete this section.
1. Use the substitution u=x2+4 to find ∫01x2+4xdx.
Hint: What is du in terms of xdx? What are the new limits?
du=2xdx, so xdx=2du. Limits: x=0⇒u=4, x=1⇒u=5.
21∫45u−1/2du=21[2u1/2]45=5−2
2. Evaluate ∫12x3+13x2dx.
Hint: What is the derivative of the denominator?
The numerator 3x2 is the derivative of the denominator x3+1. So the integral is [ln∣x3+1∣]12=ln9−ln2=ln29.
Many integrals that could be solved by substitution can be done faster by recognition. The idea is to spot that the integrand has the form:
f′(g(x))⋅g′(x)
This is the derivative of f(g(x)) by the chain rule. So:
∫f′(g(x))⋅g′(x)dx=f(g(x))+c
The most common pattern is:
∫[g(x)]n⋅g′(x)dx=n+1[g(x)]n+1+c
Ask yourself: “Is the derivative of the bracket sitting outside?” If so, you can write down the answer directly.
Find ∫3x2(x3+4)5dx.
The bracket is x3+4. Its derivative is 3x2 — which is exactly what appears outside. So by the reverse chain rule:
∫3x2(x3+4)5dx=6(x3+4)6+c
Find ∫cosx⋅esinxdx.
The inner function is sinx, and its derivative cosx appears as a factor. This is the reverse chain rule for eg(x):
∫g′(x)⋅eg(x)dx=eg(x)+c
So: ∫cosx⋅esinxdx=esinx+c.
A particularly important case:
∫f(x)f′(x)dx=ln∣f(x)∣+c
Find ∫x2+12xdx.
The denominator is x2+1. Its derivative is 2x — which is the numerator. So:
∫x2+12xdx=ln∣x2+1∣+c=ln(x2+1)+c
(We can drop the absolute value since x2+1>0 for all x.)
Exam tip: Whenever you see a fraction where the numerator is (or is a constant multiple of) the derivative of the denominator, the answer involves ln of the denominator. This pattern is tested frequently.
Answer at least 3 of 3 correctly to complete this section.
Using the substitution u=1+2x, find ∫041+2xxdx.
With u=1+2x: dxdu=2, so dx=2du.
We also need x in terms of u: from u=1+2x, we get x=2u−1.
Change the limits:
∫041+2xxdx=∫19u(u−1)/2⋅2du=41∫19u1/2u−1du
=41∫19(u1/2−u−1/2)du=41[32u3/2−2u1/2]19
=41[(32⋅27−2⋅3)−(32−2)]=41[(18−6)−(−34)]
=41(12+34)=41⋅340=310
Using the substitution u=cosx, find ∫sinxcos2xdx.
With u=cosx: dxdu=−sinx, so sinxdx=−du.
∫sinxcos2xdx=∫u2⋅(−du)=−∫u2du=−3u3+c=−3cos3x+c
Using the substitution u=x, find ∫x(1+x)1dx.
With u=x=x1/2: dxdu=2x1, so xdx=2du.
∫x(1+x)1dx=∫1+u1⋅2du=2∫1+u1du
=2ln∣1+u∣+c=2ln(1+x)+c
(Since 1+x>0, we can drop the absolute value.)
When the substitution is not given, follow these guidelines:
| What you see | Try u=… |
|---|---|
| (ax+b)n | u=ax+b |
| ax+b | u=ax+b (or u=ax+b) |
| f′(x)⋅[f(x)]n | u=f(x) (reverse chain rule) |
| f(x)f′(x) | u=f(x) (gives ln) |
| sinnxcosx or cosnxsinx | u=sinx or u=cosx |
| ef(x)⋅f′(x) | u=f(x) |
Common misconception: Students sometimes think there is a “right” substitution that you must spot instantly. In practice, if your first choice does not simplify the integral, try a different u. The exam usually provides the substitution for harder integrals.
Answer at least 3 of 3 correctly to complete this section.
Lock in what you've learned with exam-style questions and spaced repetition.
Differentiating composite functions using the chain rule.
Integration as the reverse of differentiation, integrating x^n, finding constants of integration.
Solving first-order differential equations by separation of variables, modelling with DEs.