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Techniques of integration
Integration by parts, trigonometric integrals, trigonometric substitution, partial fractions, choosing a strategy.
Lecture 9 · Tutorials 9 and 10 (first part) · Techniques of integration and improper integrals
Textbook sections: Thomas ch. 8 Techniques of Integration (8.1–8.5, 8.8) · Stewart ch. 7 (7.1–7.5, 7.8)
Textbook sections: Thomas ch. 8 Techniques of Integration (8.1–8.5, 8.8) · Stewart ch. 7 (7.1–7.5, 7.8)
Techniques of integration
Calculus I · Module 10
After this module you can
- Choose between substitution, parts, trigonometric identities, trigonometric substitution and partial fractions.
- Evaluate standard integrals fast and verify them by differentiation.
- Recognise the pattern at a glance: LIATE for parts, identities for trigonometric powers, $\sqrt{a^2\pm x^2}$ and $\sqrt{x^2-a^2}$ for trigonometric substitution.
Integration by parts
$$\int u\,dv=uv-\int v\,du$$
- Choose $u$ by LIATE: Logarithm, Inverse trig, Algebraic, Trigonometric, Exponential; $dv$ the rest.
- $\int x\cos x\,dx=x\sin x+\cos x$; $\int x^2e^xdx=(x^2-2x+2)e^x$.
- Cyclic: $\int e^x\cos x\,dx=\frac{e^x(\sin x+\cos x)}{2}$ (apply parts twice and solve for the integral).
Trigonometric integrals
| Integral | Idea |
|---|---|
| $\sin^m x\cos^n x$, $n$ odd | save one $\cos x$, $u=\sin x$ |
| $m,n$ even | half-angle: $\sin^2x=\frac{1-\cos2x}2$ |
| $\tan^m x\sec^n x$, $n$ even | save $\sec^2x$, $u=\tan x$ |
| $\sin ax\cos bx$ | product-to-sum |
Examples: $\int\cos^5x\,dx=\sin x-\frac23\sin^3x+\frac15\sin^5x$; $\int\tan^4x\,dx=\frac{\tan^3x}3-\tan x+x$.
Trigonometric substitution
| Contains | Substitution | Identity |
|---|---|---|
| $\sqrt{a^2-x^2}$ | $x=a\sin\theta$ | $1-\sin^2=\cos^2$ |
| $\sqrt{a^2+x^2}$ | $x=a\tan\theta$ | $1+\tan^2=\sec^2$ |
| $\sqrt{x^2-a^2}$ | $x=a\sec\theta$ | $\sec^2-1=\tan^2$ |
Draw the right triangle to convert back. $\int\frac{dx}{\sqrt{4+x^2}}=\ln\left(x+\sqrt{4+x^2}\right)+C$.
Partial fractions
- Proper fraction $\frac PQ$ ($\deg P<\deg Q$); else divide first: $\frac{2x^3-2x^2+1}{x^2-x}=2x+\frac1{x-1}-\frac1x$.
- Factor $Q$: linear factors give $\frac A{x-r}$ (repeated: $\frac{A_1}{x-r}+\frac{A_2}{(x-r)^2}+\cdots$); irreducible quadratics give $\frac{Bx+C}{x^2+px+q}$.
- Cover-up: $\frac{x^2+4x+1}{(x-1)(x+1)(x+3)}$: $A=\frac34$, $B=\frac12$, $C=-\frac14$.
- Integrate: logs for linear factors, $\arctan$ for quadratics.
Typical mistakes
PartsChoosing $u$ that gets harder (e.g. $u=e^x$ with $dv=x\,dx$).
Trig substitutionForgetting $dx=a\cos\theta\,d\theta$ and the back-substitution.
Partial fractionsNot dividing first when $\deg P\ge\deg Q$.
Practice next
- Chapter quizzes Ch. 5 (3 sets)
- Practice bank, filtered to this chapter, then written mode
- Lab: Antiderivative / derivative checker
- Then the mock exams (Exams page)
Worked examples from the course problems
Worked example 1 · Techniques of integration (Tutorial 9, Q1(b))
Evaluate $\displaystyle\int \dfrac{e^{-\cot z}}{\sin^2z}\,dz$.
Show the answer and the solution
Answer: $e^{- \frac{\cos{\left(z \right)}}{\sin{\left(z \right)}}}+C$
$u=-\cot z$, $du=\csc^2z\,dz$: $\int e^u\,du=e^u$. Check: differentiating gives the integrand. Answer: $e^{- \frac{\cos{\left(z \right)}}{\sin{\left(z \right)}}}+C$.
Worked example 2 · Integration by parts (Tutorial 9, Q4(b))
$\displaystyle\int_0^{\pi/2}x^3\cos2x\,dx=$
Show the answer and the solution
Answer: $\frac{3}{4} - \frac{3 \pi^{2}}{16}$
Integrate by parts three times (tabular method): $x^3\cdot\frac{\sin2x}2+3x^2\frac{\cos2x}{4}-6x\frac{\sin2x}{8}-6\frac{\cos2x}{16}$ evaluated from $0$ to $\frac\pi2$; the value is $\frac{3}{4} - \frac{3 \pi^{2}}{16}\approx-1.1006$.
Worked example 3 · Trigonometric substitution (Tutorial 10, Q1(b))
Evaluate $\displaystyle\int \sqrt{1-9t^2}\,dt$.
Show the answer and the solution
Answer: $\frac{t \sqrt{1 - 9 t^{2}}}{2} + \frac{\arcsin{\left(3 t \right)}}{6}+C$
$3t=\sin\theta$: $\frac13\int\cos^2\theta\,d\theta=\frac16(\theta+\sin\theta\cos\theta)$. Check: differentiating gives the integrand. Answer: $\frac{t \sqrt{1 - 9 t^{2}}}{2} + \frac{\arcsin{\left(3 t \right)}}{6}+C$.
Worked example 4 · Partial fractions (Computation problems 10, Q4)
Evaluate $\displaystyle\int \dfrac{x^2+4x+1}{(x-1)(x+1)(x+3)}\,dx$.
Show the answer and the solution
Answer: $\frac{3 \ln{\left(x - 1 \right)}}{4} + \frac{\ln{\left(x + 1 \right)}}{2} - \frac{\ln{\left(x + 3 \right)}}{4}+C$
Cover-up: $A=\frac{1+4+1}{2\cdot4}=\frac34$, $B=\frac{1-4+1}{(-2)(2)}=\frac12$, $C=\frac{9-12+1}{(-4)(-2)}=-\frac14$. Check: differentiating gives the integrand. Answer: $\frac{3 \ln{\left(x - 1 \right)}}{4} + \frac{\ln{\left(x + 1 \right)}}{2} - \frac{\ln{\left(x + 3 \right)}}{4}+C$.
Practice
Quiz set A Quiz set B Quiz set C Practice bank (this chapter) Written mode Lab: Antiderivative / derivative checker