3.5 Integration
- Syllabus
- 9709–2028–2029
- Topic
- 3.5
- Level
- A2
Learning objectives
3.5.1Extended integration rules• extend the idea of 'reverse differentiation' to include the integration of eax + b, ax b 1 +, sin(ax + b), cos(ax + b), sec2(ax + b) and xa 1 22+ Including examples such as x23 1 2+.3.5.2Trig integration• use trigonometrical relationships in carrying out integration e.g. use of double-angle formulae to integrate sin2 x or cos2(2x).3.5.3Partial fractions• integrate rational functions by means of decomposition into partial fractions Restricted to types of partial fractions as specified in topic 3.1 above.3.5.4An integrand of the form x• recognise an integrand of the form x k x f f l ^ ^ h h, and integrate such functions e.g. integration of x x 12 +, tan x.3.5.5Inverse tangent• recognise when an integrand can usefully be regarded as a product, and use integration by parts e.g. integration of x sin 2x, x2e-x, ln x, x tan-1 x.3.5.6Integration by substitution• use a given substitution to simplify and evaluate either a definite or an indefinite integral. e.g. to integrate sin2 2x cos x using the substitution u = sin x.
Match one of the six A2 standard antiderivatives
| Integrand ($a
| e0$) | Antiderivative |
|---|---|
| eax+b | eax+b/a+C |
| 1/(ax+b) | (1/a)ln∣ax+b∣+C |
| sin(ax+b) | −cos(ax+b)/a+C |
| cos(ax+b) | sin(ax+b)/a+C |
| sec2(ax+b) | an(ax+b)/a+C |
| 1/(x2+a2), a>0 | (1/a)tan−1(x/a)+C |
Match the whole integrand to one row, keep the linear inner expression unchanged, divide by its gradient a, carry any outside constant, and include +C for an indefinite integral.
\int 3e^{2x-1},dx= rac32e^{2x-1}+C,\int rac{5}{3x+4},dx= rac53\ln|3x+4|+C.
Differentiate the answer: the chain factor a must cancel the inserted 1/a. For definite integrals, use an interval that does not cross a point where the integrand is undefined.
These are direct reverse-derivative forms. Do not introduce substitution, integration by parts or partial fractions unless a later objective explicitly calls for them.
Reduce squared trig functions before integrating
\sin^2u= rac{1-\cos2u}{2},\qquad \cos^2u= rac{1+\cos2u}{2}.These follow from the two useful forms of $\cos2u$.
First rewrite the squared sine or cosine as a constant plus/minus a double-angle cosine. Then integrate term by term using the linear-inner rule, including the factor created by the doubled angle.
\int\sin^2x,dx=\int rac{1-\cos2x}{2},dx= rac x2- rac{\sin2x}{4}+C.
\int\cos^2(2x),dx=\int rac{1+\cos4x}{2},dx= rac x2+ rac{\sin4x}{8}+C.
The square is on the trig value, so the ordinary power integration rule does not apply. This objective uses trig identities with the direct P2 antiderivatives, not a general substitution method.
Integrate each permitted partial-fraction term by its own pattern
First decompose using only the three denominator structures approved in 3.1. Then integrate every resulting linear, repeated-linear or linear-over-quadratic term separately.
| Term | Antiderivative pattern |
|---|---|
| A/(ax+b) | (A/a)ln∣ax+b∣ |
| A/(ax+b)2 | −A/[a(ax+b)] |
| (Bx+C)/(ax2+c) | split numerator into a multiple of 2ax plus a constant; log plus possible inverse tangent |
\int\frac1{x(x+1)},dx=\int\left(\frac1x-\frac1{x+1}\right)dx=\ln|x|-\ln|x+1|+C.
Recombine the decomposition before integration, then differentiate the final antiderivative. Preserve intervals that do not cross denominator zeros.
Do not assign a constant numerator to an irreducible quadratic when a linear numerator is required, and do not integrate the original quotient by dividing numerator and denominator termwise.
Recognise a scaled logarithmic derivative
Where $f(x)\ne0$:\int k\frac{f'(x)}{f(x)},dx=k\ln|f(x)|+C.
Differentiate the denominator or inner function, compare it with the numerator, factor out the required constant, then apply the rule. If a leftover remains, split it and use another approved standard form.
\int\frac{x}{x^2+1},dx=\frac12\int\frac{2x}{x^2+1},dx=\frac12\ln(x^2+1)+C.
\int\tan x,dx=\int\frac{\sin x}{\cos x},dx=-\ln|\cos x|+C.
The numerator need only be proportional to f′, not identical. Absolute values are required unless f is known positive on the interval.
Reverse the product rule with integration by parts
\int u,dv=uv-\int v,du.Choose $u$ to simplify on differentiation and $dv$ to have a known antiderivative.
Identify the product (write lnx=1⋅lnx if needed), state u,dv,du,v, substitute into the formula, evaluate the remaining simpler integral and add C.
\int xe^x,dx=xe^x-\int e^x,dx=e^x(x-1)+C.
\int\ln x,dx=x\ln x-x+C,\qquad x>0.
The tree short label says inverse tangent, but the official objective requires integration by parts, including products such as xtan−1x. Do not swap du with v.
Apply the given substitution to every part of the integral
Use the substitution supplied by the question. Find du, rewrite every factor and dx in u, and do not mix variables. For a definite integral, convert both limits immediately or substitute back before using the original limits.
Given $u=\sin x$:\int\sin^2x\cos x,dx=\int u^2,du=\frac{u^3}{3}+C=\frac{\sin^3x}{3}+C.
For definite limits x=a,b, replace them by u(a),u(b) and finish entirely in u. If the substitution is not one-to-one over the interval, follow the question structure carefully and verify the transformed bounds.
Differentiate an indefinite answer or compare a definite result numerically/sign-wise with the original integrand.
This objective asks for use of a given substitution, not invention of a general substitution strategy. The differential factor and all limits are part of the substitution.