Use a given substitution and transform every integration limit
Rewrite the differential and all remaining factors in the new variable, change definite limits, and return to the original variable for an indefinite answer.
Exam scope and assessment notes
Pure Mathematics 3 (Paper 3, A Level) · Mathematics 9709, 2026–2027, section 3.5. Paper 1 integration, shared standard integrals and identities, Paper 3 partial fractions and derivatives are assumed. Rational decompositions follow the three denominator forms in section 3.1; higher-degree numerators are excluded, while equal-degree cases may need division first. Substitution questions use a given substitution. Trigonometric and inverse-tangent outputs use radians. Written practice is self-checked; human subject-teacher review remains pending.
Before you begin
- Differentiate composites and use double-angle identities.
- Integrate powers and exponentials and evaluate endpoints in a single variable.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
For u = sin x with radian input, what is du?
What you will learn
- Use a supplied substitution to simplify definite and indefinite integrals.
- Transform both endpoints and the entire integrand without mixing variables or reversing the orientation unnoticed.
A substitution rewrites the whole differential expression
If a problem gives u = g(x), differentiate to obtain du = g′(x) dx and use it to replace the matching differential factor. Express every remaining factor in u as well. For ∫sin²(2x)cos x dx with the given u = sin x, du = cos x dx. Since sin(2x) = 2sin x cos x and cos²x = 1 − sin²x, sin²(2x) becomes 4u²(1 − u²). The transformed integral is 4∫(u² − u⁴) du, with no x left in it.
The primitive in u is 4u³/3 − 4u⁵/5 + C. For an indefinite integral, replace u by sin x to give 4sin³x/3 − 4sin⁵x/5 + C. Differentiation returns 4sin²x cos³x = sin²(2x)cos x, so the identity and differential conversion are independently checked. The syllabus asks you to use a given substitution; choosing an unsupported substitution is not needed for this objective.
Definite limits belong to the variable of integration
On x from 0 to π/6, the substitution u = sin x changes the limits to 0 and 1/2. The transformed integral is 4∫ from 0 to 1/2 of (u² − u⁴) du. Evaluating gives 1/6 − 1/40 = 17/120. Do not keep the radian x limit π/6 on a u integral. Alternatively, back-substitute the primitive and then evaluate the original x endpoints; either method uses one consistent variable at a time.
For a given u = 1 + x², du = 2x dx, so ∫ from 0 to 1 of 2x/(1 + x²)² dx becomes ∫ from 1 to 2 of u^(−2) du = 1/2. A decreasing substitution may reverse the endpoint order; keep the derivative sign and transformed orientation together rather than discarding both independently. If an inverse expression is needed for x or a square root, use the original interval to choose its valid branch. A missing derivative factor cannot simply disappear.
Pause and explain
After substituting u = sin x in an integral from x = 0 to x = π/6, which limits go with du?
Worked example
Use the given substitution u = sin x to integrate sin²(2x)cos x, then evaluate from 0 to π/6.
Show the worked solution
- Set du = cos x dx and rewrite sin²(2x) as 4u²(1 − u²).
- Integrate to get 4u³/3 − 4u⁵/5 + C, or back-substitute for an indefinite answer.
- For the definite integral use u limits 0 and 1/2, giving 1/6 − 1/40 = 17/120.
Answer Primitive 4sin³x/3 − 4sin⁵x/5 + C; definite integral 17/120.
Try a topic test
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Using the given u = 1 + x², evaluate ∫ from 0 to 1 of 2x/(1 + x²)² dx.
Give me a hint
The numerator times dx is exactly du; transform both bounds.
Compare my reasoning
- Since du = 2x dx, the integrand becomes u^(−2) du.
- The bounds become u = 1 and u = 2.
- The primitive −1/u gives −1/2 − (−1) = 1/2.
1/2.
Look for these in your work
- Transform the differential and both limits.
- Apply the negative-power rule with the correct sign.
Use the given u = x² + 3 to evaluate ∫ from 1 to 2 of x√(x² + 3) dx, and give its primitive in x.
Give me a hint
Use x dx = du/2 and change bounds to the values of x² + 3.
Compare my reasoning
- The transformed definite integral is (1/2)∫ from 4 to 7 of u^(1/2) du.
- Its primitive is u^(3/2)/3, so the result is (7√7 − 8)/3.
- Back-substitution gives primitive (x² + 3)^(3/2)/3 + C, whose derivative is x√(x² + 3).
Definite integral (7√7 − 8)/3; primitive (x² + 3)^(3/2)/3 + C.
Look for these in your work
- Retain the factor 1/2 from the differential.
- Use one variable at a time and verify the back-substituted primitive.
A vessel starts with 4 litres and has supplied inflow q(t) = t e^(t²) litres per minute for 0 ≤ t ≤ √(ln 2), with t measured numerically in minutes. Use the given u = t² to find added volume and final total.
Give me a hint
The differential t dt is du/2; the final transformed endpoint is ln 2.
Compare my reasoning
- With du = 2t dt, the accumulated volume is (1/2)∫ from 0 to ln 2 of eᵘ du.
- Evaluation gives (1/2)(2 − 1) = 1/2 litre added.
- Adding the original 4 litres gives final total 9/2 = 4.5 litres.
Added volume 1/2 litre; final total 9/2 = 4.5 litres.
Look for these in your work
- Include the differential factor and transform the stated endpoint.
- Interpret the rate's accumulation before adding the initial stock.
Common mistakes
- Rewrite every factor and the differential, rather than only renaming the denominator.
- Use u endpoints for a u integral, or back-substitute before using x endpoints.
- For an indefinite answer return to the original variable and keep C.
What can you explain now?
With u = 1 + x², what are the transformed limits for x from 0 to 1?
Compare with the explanation
u from 1 to 2.
Evaluate the given substitution at both original endpoints.
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Make it stick
For a given substitution, record its derivative, rewrite every factor and transform both definite limits. Explain why mixed variables or mixed limits cannot be used in the same integral.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.