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AI HL · AHL 3

Further Integration

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AHL 3.2

Integrating f(ax+b) and by inspection

AHL 3.2

Integrating f(ax+b) and by inspection

Target: I can integrate \((ax+b)^n\), \(\sin(ax+b)\), \(\cos(ax+b)\), \(e^{ax+b}\) and \(\frac{1}{ax+b}\) by dividing by \(a\), including when \(a\) is negative. (+5 more)

By the end of this section

Key points

A linear expression inside

  • If \(\int f(x)\,dx=F(x)+c\), then \(\int f(ax+b)\,dx=\frac1a F(ax+b)+c\). Integrate as normal, then divide by \(a\).
  • \(\int(ax+b)^n\,dx=\frac{(ax+b)^{n+1}}{a(n+1)}+c\) for \(n\neq-1\). Roots and fractions become powers first.
  • \(\int\cos(ax+b)\,dx=\frac1a\sin(ax+b)+c\) and \(\int\sin(ax+b)\,dx=-\frac1a\cos(ax+b)+c\).
  • \(\int e^{ax+b}\,dx=\frac1a e^{ax+b}+c\) and \(\int\frac{1}{ax+b}\,dx=\frac1a\ln|ax+b|+c\).
  • Divide by \(a\) with its sign. For \(\int\frac{1}{3-2x}\,dx\) you divide by \(-2\), so the answer is \(-\frac12\ln|3-2x|+c\).

Why you divide by a

  • Let \(u=ax+b\). Then \(\frac{du}{dx}=a\), so \(dx=\frac{du}{a}\).
  • Example: \(\int\cos(2x+7)\,dx\) with \(u=2x+7\), \(dx=\frac{du}{2}\), becomes \(\int\cos u\cdot\frac{du}{2}=\frac12\sin u+c=\frac12\sin(2x+7)+c\).
  • Differentiating \(\frac12\sin(2x+7)\) with the chain rule multiplies by 2 and cancels the \(\frac12\). That is the check.

Reversing the chain rule by inspection

  • \(\int g'(x)\,f'(g(x))\,dx=f(g(x))+c\). Look for a function \(g(x)\) and its derivative \(g'(x)\) in the same integrand.
  • Three patterns: \(\int g'(x)[g(x)]^n\,dx=\frac{[g(x)]^{n+1}}{n+1}+c\) for \(n\neq-1\); \(\int\frac{g'(x)}{g(x)}\,dx=\ln|g(x)|+c\); \(\int g'(x)e^{g(x)}\,dx=e^{g(x)}+c\).
  • Trig cases: \(\int\cos x\sin^n x\,dx=\frac{\sin^{n+1}x}{n+1}+c\) and \(\int\sin x\cos^n x\,dx=-\frac{\cos^{n+1}x}{n+1}+c\).
  • If the integrand is a constant times \(g'\), integrate and keep the constant: \(\int 6x(x^2+3)^5\,dx=3\int 2x(x^2+3)^5\,dx\). A factor of x cannot be fixed this way.
  • Always check by differentiating. If the result is not the integrand, the pattern was wrong.

One over cos squared

  • \(\frac{d}{dx}\tan x=\frac{1}{\cos^2 x}\), so \(\int\frac{1}{\cos^2 x}\,dx=\tan x+c\). There is no minus sign: the minus belongs to the integral of \(\sin x\).
  • With a linear expression inside, divide by a: \(\int\frac{1}{\cos^2(ax+b)}\,dx=\frac1a\tan(ax+b)+c\).

Worked examples

Problems like the ones in class. Pick one, try it, then open the working.
A cube root of a linear expression

Find \(\int \sqrt[3]{6x+1}\,dx\). Give the coefficient of the bracket as a decimal.

Working
  1. Write as a power

    \(\sqrt[3]{6x+1}=(6x+1)^{\frac13}\)

    Rewrite the root first.

  2. Integrate as normal

    \(\frac13+1=\frac43,\quad \frac{(6x+1)^{4/3}}{4/3}\)

    Treat 6x + 1 as if it were x.

  3. Divide by 6

    \(\frac34\div 6=\frac{1}{8}\)

    The 6 is the coefficient of x. The chain rule would have multiplied by 6, so divide.

  4. Answer

    \(\int \sqrt[3]{6x+1}\,dx=\frac{1}{8}(6x+1)^{\frac{4}{3}}+c\)

    Add c.

Video walkthrough coming
A negative power of a linear expression

Find \(\int 6(2x-1)^{-\frac{3}{2}}\,dx\).

Working
  1. Add 1

    \(-\frac32+1=-\frac12\)

    New power: -1/2.

  2. Divide by the new power

    \(6\div\left(-\frac12\right)=-12\)

    Dividing by -1/2 gives -12.

  3. Divide by 2

    \(-12\div 2=-6\)

    2 is the coefficient of x.

  4. Answer

    \(\int 6(2x-1)^{-\frac{3}{2}}\,dx=-6(2x-1)^{-\frac{1}{2}}+c\)

    Also written \(-\frac{6}{\sqrt{2x-1}}+c\).

Video walkthrough coming
A sine with a negative coefficient of x

Find \(\int \sin(3-2x)\,dx\). Give the coefficient of \(\cos(3-2x)\).

Working
  1. Integrate sin

    \(\int\sin u\,du=-\cos u\)

    The integral of sin is minus cos.

  2. Divide by -2

    \(\frac{-1}{-2}=\frac12\)

    Divide by the coefficient of x, which is -2.

  3. Answer

    \(\int \sin(3-2x)\,dx=\frac{1}{2}\cos(3-2x)+c\)

    Two minus signs make a plus.

Video walkthrough coming
An exponential with a linear power

Find \(\int 6e^{2-3x}\,dx\). Give the coefficient of \(e^{2-3x}\).

Working
  1. Integrate

    \(\int 6e^u\,du=6e^u\)

    Treat the power as if it were x.

  2. Divide by -3

    \(6\div(-3)=-2\)

    Divide by the coefficient of x.

  3. Answer

    \(\int 6e^{2-3x}\,dx=-2e^{2-3x}+c\)

    The power stays as it was, 2 - 3x.

Video walkthrough coming
One over a linear expression

Find \(\int \frac{1}{5-4x}\,dx\). Give the coefficient of \(\ln|5-4x|\).

Working
  1. Integrate

    \(\int\frac1u\,du=\ln|u|\)

    The integral of 1/x is ln|x|.

  2. Divide by -4

    \(1\div(-4)=-\frac14\)

    Divide by the coefficient of x.

  3. Answer

    \(\int \frac{1}{5-4x}\,dx=-\frac{1}{4}\ln|5-4x|+c\)

    Keep the modulus signs.

Video walkthrough coming
A power of a function, by inspection

Find \(\int 6x(x^2+3)^{5}\,dx\). Give the coefficient of \((x^2+3)^6\).

Working
  1. Spot g and g'

    \(g=x^2+3,\quad g'=2x\)

    The 6x on top is a multiple of g'.

  2. Fix the multiple

    \(6x=3\times 2x=3g'\)

    Only a constant factor can be fixed this way.

  3. Integrate

    \(3\times\frac{(x^2+3)^6}{6}=\frac12(x^2+3)^6\)

    Add 1 to the power, divide by 6.

  4. Check by substitution

    \(u=x^2+3,\ du=2x\,dx,\ 6x\,dx=3\,du:\ \int3u^5\,du=\frac12u^6\)

    Same answer. Inspection is substitution done in your head.

Video walkthrough coming
A power of cos, with sin beside it

Find \(\int \sin x\cos^{3}x\,dx\). Give the coefficient of \(\cos^4 x\).

Working
  1. Spot the pair

    \(g=\cos x,\quad g'=-\sin x\)

    The sin x on the outside is minus g'.

  2. Allow for the sign

    \(\sin x\cos^3x=-g'g^3\)

    A minus sign is a constant factor, so it is allowed.

  3. Integrate

    \(-\frac{\cos^4x}{4}\)

    Add 1 to the power, divide by 4.

  4. Answer

    \(\int \sin x\cos^{3}x\,dx=-\frac{1}{4}\cos^{4}x+c\)

    Differentiate it to check: the chain rule brings back \(\sin x\cos^3x\).

Video walkthrough coming
The top is the derivative of the bottom

Find \(\int \frac{6x^{2}}{x^3+4}\,dx\). Give the coefficient of \(\ln|x^3+4|\).

Working
  1. Spot g and g'

    \(g=x^3+4,\quad g'=3x^2\)

    Differentiate the bottom.

  2. Compare with the top

    \(6x^2=2\times3x^2=2g'\)

    The top is twice g'.

  3. Integrate

    \(\int\frac{2g'}{g}\,dx=2\ln|g|\)

    g'/g integrates to ln|g|.

  4. Answer

    \(\int \frac{6x^{2}}{x^3+4}\,dx=2\ln|x^3+4|+c\)

    Keep the modulus signs.

Video walkthrough coming
e to a function, times its derivative

Find \(\int 4\sin x\,e^{\cos x}\,dx\). Give the coefficient of \(e^{\cos x}\).

Working
  1. Spot g and g'

    \(g=\cos x,\quad g'=-\sin x\)

    The sin x is minus g'.

  2. Allow for the sign

    \(4\sin x\,e^{\cos x}=-4g'e^{g}\)

    Pull out -4.

  3. Integrate

    \(\int-4g'e^g\,dx=-4e^g\)

    \(e^g\) times \(g'\) integrates to \(e^g\).

  4. Answer

    \(\int 4\sin x\,e^{\cos x}\,dx=-4e^{\cos x}+c\)

    Differentiate to check.

Video walkthrough coming
One over cos squared

Find \(\int \frac{10}{\cos^2(4x+1)}\,dx\). Give the coefficient of \(\tan(4x+1)\).

Working
  1. The pair to know

    \(\frac{d}{dx}\tan x=\frac{1}{\cos^2 x}\Rightarrow\int\frac{1}{\cos^2 u}\,du=\tan u\)

    The integral of 1 over cos squared is tan. No minus sign.

  2. Divide by 4

    \(\int\frac{10}{\cos^2(4x+1)}\,dx=10\times\frac14\tan(4x+1)\)

    4 is the coefficient of x.

  3. Answer

    \(=2.5\tan(4x+1)+c\)

    Add c.

Video walkthrough coming