Evaluate ∫14 1/(x2 − 2x + 10) dx, giving an exact answer.
- Complete the square: x2 − 2x + 10 = (x − 1)2 + 9 = (x − 1)2 + 32.
- Integrate: (1/3) tan−1((x − 1)/3).
- Put in the limits: (1/3)[tan−1(3/3) − tan−1(0)] = (1/3)(π/4 − 0).
Answer: π/12
Integration
Further Maths · Further Pure Mathematics 2. The idea in plain words, the rules to remember, the mistake to avoid and a worked example, written and checked by Mega Lecture.
The idea
Three standard forms give inverse functions when integrated: 1/√(a2 − x2), 1/(x2 + a2) and 1/√(x2 ± a2). If the quadratic has an x term, complete the square first.
A reduction formula links an integral In to one with a smaller n, usually by integration by parts. Use it again and again until you reach an integral you can do directly.
Integration also gives the arc length of a curve and the area of the surface made when a curve is rotated about an axis. Rectangles drawn under and over a curve compare a sum with an integral, which gives bounds and limits for sums.
Rules to remember
- ∫ sinh x dx = cosh x + c; ∫ cosh x dx = sinh x + c; ∫ sech2 x dx = tanh x + c
- ∫ 1/√(a2 − x2) dx = sin−1(x/a) + c; ∫ 1/(x2 + a2) dx = (1/a) tan−1(x/a) + c
- ∫ 1/√(x2 + a2) dx = sinh−1(x/a) + c; ∫ 1/√(x2 − a2) dx = cosh−1(x/a) + c for x > a
- Arc length: ∫ √(1 + (dy/dx)2) dx, or ∫ √((dx/dt)2 + (dy/dt)2) dt, or in polars ∫ √(r2 + (dr/dθ)2) dθ
- Surface area about the x-axis: 2π ∫ y √(1 + (dy/dx)2) dx; about the y-axis: 2π ∫ x √(1 + (dx/dy)2) dy
- Rectangles of unit width, f decreasing: f(2) + f(3) + … + f(n) < ∫1n f(x) dx < f(1) + f(2) + … + f(n − 1)
- Limit of a sum: as n → ∞, (1/n) Σr=1n f(r/n) → ∫01 f(x) dx
Common mistake
Students forget the factor 1/a in ∫ 1/(x2 + a2) dx = (1/a) tan−1(x/a), or put a 1/a in front of sin−1(x/a) where it does not belong. Only the tan−1 form has the 1/a.
Worked example
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