01 · Recover a particular curve
dy/dx = 2x and y = −1 when x = 1. Find y.
Hint
Substitute the point into x² + C.
Worked solution
−1 = 1 + C gives C = −2, so y = x² − 2.
Understand · explore · practise
Standard integrals
Learn A-level standard integrals of powers, exponentials, logarithms and trigonometric functions. Explore antiderivative families, check signs and practise exact definite integrals.
Before you startPower, exponential and trigonometric differentiation; Pure 1 integration.
01 / Integration reverses differentiation
An antiderivative has the required derivative.
To find ∫f(x) dx, look for a function F whose derivative is f. On an interval, every antiderivative differs from F by a constant: ∫f(x) dx = F(x) + C.
Change C in the model, keeping x fixed. The point and tangent move up or down together; the tangent gradient stays equal to f(x). The constant also cancels from differences F(b) − F(a).
Same derivative, different constantExplore
F(1) = 1 for F(x) = x².
F′(1) = 2, equal to the integrand at x = 1.
F(2) − F(0.5) = 3.75, independent of C.
All curves use the same scales. The model shows x from 0.25 to 2; the reciprocal example is on its positive domain. Trig inputs are in radians.
02 / Why the constant matters
A derivative does not record vertical position.
∫2x dx = x² + C
Differentiating x² + C gives 2x for every constant C.
A curve has dy/dx = 2x and passes through (2, 7).Worked example
y = x² + C
Start with the whole family.
7 = 4 + C, so C = 3
The point chooses one member.
y = x² + 3
Differentiate and check the original point.
Watch: shift the curve while its gradient stays fixed
Pause, replay or seek freely. The notes explain the same idea and stay in view.
03 / Integrate powers
Increase the exponent by one, then divide.
∫xⁿ dx = xⁿ⁺¹/(n + 1) + C, n ≠ −1
Work on an interval where the real-valued power and its derivative are defined.
Integrate 6x² − 3x⁻² + 2√x, for x > 0.Worked example
Rewrite √x as x¹ᐟ²
Use powers with the correct exponents.
∫6x² dx = 2x³
Divide by the new exponent 3.
∫−3x⁻² dx = 3x⁻¹
The new exponent is −1.
∫2x¹ᐟ² dx = (4/3)x³ᐟ²
Dividing by 3/2 multiplies by 2/3.
2x³ + 3/x + (4/3)x³ᐟ² + C
Use one arbitrary constant for the sum.
02 · Fractional power
Find ∫5/√x dx, for x > 0.
Hint
The exponent is −1/2.
Worked solution
10√x + C. Its derivative is 5/√x.
03 · Negative power
Find ∫4x⁻³ dx, for x ≠ 0.
Hint
The new exponent is −2.
Worked solution
−2x⁻² + C, on either connected interval avoiding zero.
04 / The reciprocal exception
The exponent −1 produces a logarithm.
∫1/x dx = ln|x| + C
Use a connected interval with x ≠ 0. For x > 0 this is ln x + C; for x < 0 it is ln(−x) + C.
Putting n = −1 in the power formula would divide by zero. It does not say the integral is zero or undefined: the antiderivative is a different type of function. Constants on disconnected positive and negative intervals need not agree.
04 · A negative interval
Find ∫7/x dx on x < 0.
Hint
Use a positive logarithm argument.
Worked solution
7ln(−x) + C, equivalently 7ln|x| + C. Differentiating gives 7/x.
05 · A false power rule
A student writes ∫x⁻¹ dx = x⁰ + C. Explain the failure.
Hint
Differentiate their answer.
Worked solution
The derivative is zero, not 1/x. The power formula excludes n = −1; the correct result is ln|x| + C.
05 / Integrate the natural exponential
The derivative of eˣ is eˣ.
∫eˣ dx = eˣ + C
Find ∫(3eˣ − 4) dx.Worked example
∫3eˣ dx = 3eˣ
Keep the constant multiplier.
∫−4 dx = −4x
A constant integrand produces a linear term.
3eˣ − 4x + C
Differentiate term by term to check.
This rule is for eˣ. For eᵃˣ with nonzero a, a chain-rule factor is needed; the next lesson handles that. Do not silently treat every exponential as eˣ.
06 · Exponential and reciprocal
Find ∫(2eˣ + 3/x) dx for x > 0.
Hint
Integrate the two terms independently.
Worked solution
2eˣ + 3ln x + C.
06 / Integrate sine and cosine
Check the sign by differentiating.
∫cos x dx = sin x + C
∫sin x dx = −cos x + C
These standard calculus identities use radians.
Find ∫(4sin x − 3cos x) dx.Worked example
4∫sin x dx = −4cos x
The minus sign is essential.
−3∫cos x dx = −3sin x
Keep the original negative coefficient.
−4cos x − 3sin x + C
Derivative: 4sin x − 3cos x.
07 · Sign check
Find ∫(2cos x + 5sin x) dx.
Hint
Differentiate your proposed answer.
Worked solution
2sin x − 5cos x + C.
08 · Square versus double angle
Can ∫sin²x dx be found by writing −cos²x?
Hint
Differentiate −cos²x.
Worked solution
No: its derivative is 2sin x cos x, not sin²x. A trigonometric identity is needed; a later lesson covers this.
07 / Integrate sec² and cosec²
Recognise the derivatives of tan and cot.
∫sec²x dx = tan x + C
∫cosec²x dx = −cot x + C
Choose an interval that excludes the poles of the integrand.
Find ∫(3sec²x − 2cosec²x) dx.Worked example
3∫sec²x dx = 3tan x
The derivative of tan is sec squared.
−2∫cosec²x dx = 2cot x
Two negative signs give a positive.
3tan x + 2cot x + C
Valid on an interval avoiding zeros of sin and cos.
09 · Reciprocal trig signs
Find ∫(2sec²x + 4cosec²x) dx.
Hint
The cotangent term needs a minus.
Worked solution
2tan x − 4cot x + C, away from the relevant poles.
08 / Two standard trig products
A product can match a known derivative.
∫sec x tan x dx = sec x + C
∫cosec x cot x dx = −cosec x + C
These are recognisable derivative patterns. There is no general rule saying the integral of a product is the product of the integrals.
10 · A coefficient with a sign
Find ∫−6cosec x cot x dx.
Hint
The derivative of cosec x is negative.
Worked solution
6cosec x + C, on a valid interval.
11 · Product trap
Is ∫x eˣ dx equal to (x²/2)eˣ + C?
Hint
Use the product rule to check.
Worked solution
No. Its derivative is xeˣ + (x²/2)eˣ, with an unwanted extra term. Integration by parts, introduced later, handles this product.
09 / Evaluate a definite integral
Subtract lower-bound value from upper-bound value.
∫ₐᵇ f(x) dx = F(b) − F(a)
Use an antiderivative on the whole interval; here f is continuous there.
Evaluate ∫₀^(π/2) (3cos x + 2sin x) dx.Worked example
F(x) = 3sin x − 2cos x
Both standard integrals are immediate.
F(π/2) = 3 and F(0) = −2
Use exact radian values.
The integral is 3 − (−2) = 5
Constants cancel, so the final definite value has no +C.
12 · Exact exponential and log
Evaluate ∫₁² (eˣ + 2/x) dx.
Hint
Use eˣ + 2ln x.
Worked solution
e² − e + 2ln2.
13 · Reversed limits
Evaluate ∫₂¹ 3x² dx.
Hint
Upper value minus lower value still applies.
Worked solution
[x³]₂¹ = 1 − 8 = −7. Reversing the limits reverses the sign.
10 / Check the whole interval
A formula cannot justify crossing a singularity.
For ∫₋₁¹ 1/x dx, the integrand is undefined at zero. Simply subtracting ln|1| − ln|−1| is invalid: there is no continuous integrand or antiderivative across the full interval. The ordinary improper integral diverges; a symmetric cancellation is a different concept.
Likewise, keep sec and tan expressions away from cos x = 0, and cosec and cot expressions away from sin x = 0. Domain checks are part of the method.
14 · Logarithm on the negative side
Evaluate ∫₋₄⁻₂ 1/x dx.
Hint
This interval does not cross zero.
Worked solution
ln2 − ln4 = −ln2. A negative signed integral is consistent with a negative integrand.
15 · A pole inside the bounds
Can you evaluate ∫₀^π sec²x dx by tanπ − tan0?
Hint
Check cos x inside the interval.
Worked solution
No. cos(π/2) = 0 creates a pole inside it. The ordinary improper integral diverges; the endpoint subtraction is not valid.
11 / Choose, integrate and check
Differentiate an indefinite answer before accepting it.
- Rewrite roots and reciprocals as powers when useful.
- Keep n = −1 separate.
- Recognise exp and the six trig patterns.
- Integrate sums term by term; products need their own justification.
- Include +C for an indefinite integral.
- Check domains and use F(b) − F(a) for valid definite integrals.
16 · Diagnose and repair
A proposed antiderivative of 2eˣ − 3sin x + 4/x is 2eˣ − 3cos x + 4ln|x|. Repair it.
Hint
Differentiate the cosine term.
Worked solution
2eˣ + 3cos x + 4ln|x| + C. The cosine sign was wrong and the arbitrary constant was missing. Work on a connected interval with x ≠ 0.
Section 1 of 11 · Integration reverses differentiation