Everything from the 6 lessons of this topic on a few pages: the key idea of each lesson, the diagram to remember, the points to learn and the traps examiners set. Read it, cover it, test yourself.
1Integrating exponentials, 1/(ax + b) and trigonometric functions
Syllabus 3.5
If \(\dfrac{d}{dx}F(x) = f(x)\), then \(\displaystyle\int f(ax+b)\,dx = \dfrac{1}{a}F(ax+b) + c\). Integrate as if the bracket were just \(x\), then divide by \(a\), the coefficient of \(x\). Angles are always in radians.
Fig. 1: the area under y = 1/(2x + 1) from x = 0 to x = 4 is the integral of 1/(2x + 1), which is (1/2) ln(2x + 1); between the limits this gives (1/2) ln 9 = ln 3, about 1.10 square units
Exact answers: tidy logs with the log laws and keep \(\pi\), \(e\) and surds.
Watch out!
\(\displaystyle\int \dfrac{1}{2x+1}\,dx\) is NOT \(\ln|2x+1|\). Differentiate \(\ln|2x+1|\) and you get \(\dfrac{2}{2x+1}\), twice too big, so the answer needs \(\dfrac{1}{2}\) in front. Equally, never multiply by \(a\): that is differentiating, not integrating.
\(\displaystyle\int \sin x\,dx = -\cos x\), with a MINUS. The most common lost mark is writing \(+\cos x\). A quick check: differentiate your answer and see whether you get back exactly what you started with.
Do not forget the \(\dfrac{1}{a}\) in front of \(\tan^{-1}\), and do not leave \(\tan^{-1}1\) in degrees: \(\tan^{-1}1 = \dfrac{\pi}{4}\), never 45, inside a calculus answer.
Squares of sine and cosine are integrated by lowering the power with the double-angle formula: \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\) and \(\cos^2 x = \dfrac{1}{2}(1 + \cos 2x)\). For \(\tan^2 x\) use \(\tan^2 x = \sec^2 x - 1\).
Fig. 1: y = sin²x = 1/2 - (1/2)cos 2x oscillates about the line y = 1/2; over 0 to π the cosine part integrates to zero, so the area is (1/2) × π = π/2
Worked example: Integrate sine squared
Find \(\displaystyle\int \sin^2 x\,dx\).
Use \(\cos 2x = 1 - 2\sin^2 x\), so \(\sin^2 x = \dfrac{1}{2}(1 - \cos 2x)\).
Squares of sin or cos: use \(\cos 2\theta\) to lower the power; the angle doubles.
\(\tan^2\): replace by \(\sec^2 - 1\).
Products: \(\sin x\cos x = \dfrac{1}{2}\sin 2x\).
Brackets: expand, then use \(\sin^2 + \cos^2 = 1\).
Evaluate in radians and keep \(\pi\) and surds exact.
Watch out!
\(\displaystyle\int \sin^2 x\,dx\) is NOT \(\dfrac{\sin^3 x}{3}\). The power rule only works on \(x^{n}\); differentiating \(\dfrac{\sin^3 x}{3}\) gives \(\sin^2 x\cos x\), not \(\sin^2 x\). And when the angle is \(2x\), the identity brings in \(\cos 4x\), not \(\cos 2x\).
\(\tan^2 x = \sec^2 x - 1\), not \(\sec^2 x + 1\). Check with \(x = 0\): \(\tan^2 0 = 0\) and \(\sec^2 0 - 1 = 0\). The wrong sign turns \(1 - \dfrac{\pi}{4}\) into \(1 + \dfrac{\pi}{4}\).
When you square \(\dfrac{1 + \cos 2x}{2}\), square the 2 in the denominator as well: the factor in front is \(\dfrac{1}{4}\), not \(\dfrac{1}{2}\).
3Integration using partial fractions and f'(x)/f(x)
Syllabus 3.5
\(\displaystyle\int \dfrac{f'(x)}{f(x)}\,dx = \ln|f(x)| + c\). For any other proper rational function, write it in partial fractions; then \(\dfrac{A}{ax+b} \to \dfrac{A}{a}\ln|ax+b|\), \(\dfrac{B}{(ax+b)^2} \to -\dfrac{B}{a(ax+b)}\), and \(\dfrac{Cx+D}{x^2+k^2}\) splits into a log and an inverse tan.
Fig. 1: y = (5x + 3)/((x + 1)(2x + 1)) split into partial fractions as 2/(x + 1) + 1/(2x + 1); each piece integrates to a logarithm, so the shaded area from x = 0 to x = 1 is 2 ln 2 + (1/2) ln 3, about 1.94
Worked example: Fix the constant
Find the exact value of \(\displaystyle\int_0^2 \dfrac{x}{x^2+4}\,dx\).
The derivative of \(x^2 + 4\) is \(2x\). The top is \(x = \dfrac{1}{2}(2x)\).
So \(\displaystyle\int \dfrac{x}{x^2+4}\,dx = \dfrac{1}{2}\ln(x^2+4)\) (no modulus: \(x^2 + 4 > 0\)).
Check first: if the numerator is \(k f'(x)\), the answer is \(k\ln|f(x)|\).
Improper? If the top has the same degree as the bottom, start with a constant term.
Split into partial fractions; substitute roots, then compare coefficients.
Integrate each piece: \(\ln\) for linear, power rule for squared, \(\ln\) plus \(\tan^{-1}\) for \(x^2 + k^2\).
Tidy with the log laws; give exact values unless asked otherwise.
Watch out!
\(\displaystyle\int \dfrac{x}{x^2+4}\,dx\) is not \(\ln(x^2+4)\): that differentiates to \(\dfrac{2x}{x^2+4}\). Always differentiate your log mentally and compare the numerator, then fix the constant in front.
\(\dfrac{1}{2x+1}\) integrates to \(\dfrac{1}{2}\ln|2x+1|\), not \(\ln|2x+1|\). Partial fractions with a coefficient of \(x\) other than 1 are where this mark is most often lost.
\(\dfrac{1}{(x+1)^2}\) does NOT integrate to a logarithm. Write it as \((x+1)^{-2}\) and use the power rule: \(-\dfrac{1}{x+1}\). Likewise \(\dfrac{1}{x^2+1}\) gives \(\tan^{-1}x\), not \(\ln(x^2+1)\).
With a substitution \(u = g(x)\), replace three things: the integrand (in terms of \(u\)), \(dx\) (using \(\dfrac{du}{dx}\)), and, for a definite integral, the limits (\(x\)-values become \(u\)-values). Then the whole integral is in \(u\) and no \(x\) may remain.
Fig. 1: with u = sin x, the x-limits 0, π/6 and π/2 map to the u-limits 0, 1/2 and 1 (drawn to scale on each line); a definite integral in u uses the new limits
Worked example: A linear substitution
Use the substitution \(u = 2x + 1\) to find the exact value of \(\displaystyle\int_0^4 x\sqrt{2x+1}\,dx\).
\(\dfrac{du}{dx} = 2\), so \(dx = \dfrac{1}{2}\,du\); and \(x = \dfrac{u-1}{2}\).
Limits: \(x = 0 \Rightarrow u = 1\); \(x = 4 \Rightarrow u = 9\).
The integral becomes \(\displaystyle\int_1^9 \dfrac{u-1}{2}\,u^{1/2}\times\dfrac{1}{2}\,du = \dfrac{1}{4}\int_1^9 \left(u^{3/2} - u^{1/2}\right) du\).
Differentiate the substitution and express \(dx\) (or a group like \(\cos x\,dx\)) in \(du\).
Rewrite any leftover \(x\) in terms of \(u\).
Change the limits for a definite integral.
Integrate in \(u\); simplify powers first.
Back-substitute for an indefinite integral.
Watch out!
Forgetting to replace \(dx\) is the most expensive slip: with \(u = 2x + 1\), \(dx = \dfrac{1}{2}du\), not \(du\). The second is mixing old and new limits: once the integral is in \(u\), the limits must be \(u\)-values.
With \(u = \cos x\), \(du = -\sin x\,dx\). The minus sign is easy to drop; it is usually absorbed by swapping the limits, because \(x = 0\) gives \(u = 1\), the larger value.
An indefinite answer left in \(u\) loses the final mark. A definite answer evaluated with the old \(x\)-limits in a \(u\)-integral is simply wrong.
\(\displaystyle\int u\dfrac{dv}{dx}\,dx = uv - \int v\dfrac{du}{dx}\,dx\). Choose \(u\) to be the factor that gets simpler when differentiated (a power of \(x\), or \(\ln x\) and \(\tan^{-1}x\), which you cannot integrate directly), and \(\dfrac{dv}{dx}\) to be the factor you can integrate.
Fig. 1: the area under y = ln x from x = 1 to x = e is found by parts with u = ln x and dv/dx = 1: [x ln x - x] from 1 to e = (e - e) - (0 - 1) = 1 square unit
Worked example: Power times exponential
Find \(\displaystyle\int x e^{2x}\,dx\).
\(u = x\), so \(\dfrac{du}{dx} = 1\). \(\dfrac{dv}{dx} = e^{2x}\), so \(v = \dfrac{1}{2}e^{2x}\).
u is \(\ln x\) or \(\tan^{-1}x\) if present, otherwise the power of \(x\).
Lone ln x: take \(\dfrac{dv}{dx} = 1\).
Twice for \(x^2\) times an exponential or trig function; bracket the second result.
Check by differentiating your answer.
Watch out!
Choosing \(u = e^{2x}\) and \(\dfrac{dv}{dx} = x\) makes things worse: the new integral contains \(\dfrac{x^2}{2}e^{2x}\). If the integral you produce is harder than the original, swap \(u\) and \(\dfrac{dv}{dx}\).
With \(\ln x\) never choose \(\dfrac{dv}{dx} = \ln x\): you would need to integrate \(\ln x\) to find \(v\), which is the very thing you cannot yet do.
The minus sign in front of the second integral applies to EVERYTHING it produces. When parts is used twice, bracket the second result: \(-x^2e^{-x} + \left(-2xe^{-x} - 2e^{-x}\right)\). Sign slips here cost the accuracy marks.
With \(n\) intervals of width \(h = \dfrac{b-a}{n}\) and ordinates \(y_0, y_1, \ldots, y_n\): \(\displaystyle\int_a^b y\,dx \approx \dfrac{h}{2}\left[y_0 + 2(y_1 + \cdots + y_{n-1}) + y_n\right]\). If the curve bends upwards (lies below its chords) the rule over-estimates; if it bends downwards it under-estimates.
Fig. 1: three trapezia of width h = 1 under y = x² + 1 from x = 0 to x = 3; the curve bends upwards, so every chord lies above it and the estimate 12.5 is an over-estimate of the exact area 12
Worked example: A table of values
Use the trapezium rule with 4 intervals to estimate \(\displaystyle\int_0^2 \sqrt{1+x^3}\,dx\), giving your answer to 3 significant figures.
Width: \(h = \dfrac{b-a}{n}\) with \(n\) intervals.
Table: \(n + 1\) ordinates to 4 or 5 significant figures; radians for trig.
Weights: ends once, middles twice, times \(\dfrac{h}{2}\).
Sketch: bends up gives an over-estimate, bends down an under-estimate.
Round only at the end, to the accuracy asked.
Watch out!
\(n\) intervals have \(n + 1\) ordinates. With 4 intervals on \([0, 2]\), \(h = 0.5\), not \(\dfrac{2}{5}\). Equally, do not forget the \(\dfrac{h}{2}\) at the end, or the answer comes out twice as big.
Do not judge by whether the curve is going up or down: \(y = \ln x\) is increasing, but it bends downwards, so the rule under-estimates. It is the bending (concavity) that matters, not the slope.
In degree mode \(\sin(0.785\ldots)\) is about 0.0137, not 0.707, and the estimate collapses. If a trigonometric estimate looks tiny, check the calculator mode.