Integrating f(ax+b) and by inspection Problem set B Worked solutions

Find each integral.

1. (a) \(\int \sqrt{3x+7}\,dx\)

   • Integrate as normal, treating the inside as x (add 1 to the power and divide by the new power, 3/2): \(\frac{2}{3}(3x+7)^{\frac{3}{2}}\).
   • Divide by 3, the coefficient of x: \(\frac{2}{3}\div3=\frac{2}{9}\).
   • \(\int \sqrt{3x+7}\,dx=\frac{2}{9}(3x+7)^{\frac{3}{2}}+c\)

2. (b) \(\int \sqrt[3]{2x-5}\,dx\)

   • Integrate as normal, treating the inside as x (add 1 to the power and divide by the new power, 4/3): \(\frac{3}{4}(2x-5)^{\frac{4}{3}}\).
   • Divide by 2, the coefficient of x: \(\frac{3}{4}\div2=\frac{3}{8}\).
   • \(\int \sqrt[3]{2x-5}\,dx=\frac{3}{8}(2x-5)^{\frac{4}{3}}+c\)

Find \(\int 10(5x-2)^{-\frac{3}{2}}\,dx\).

• Integrate as normal, treating the inside as x (add 1 to the power and divide by the new power, -1/2): \(-20(5x-2)^{-\frac{1}{2}}\).
• Divide by 5, the coefficient of x: \(-20\div5=-4\).
• \(\int 10(5x-2)^{-\frac{3}{2}}\,dx=-4(5x-2)^{-\frac{1}{2}}+c\)

Find each integral.

1. (a) \(\int \cos(6x+2)\,dx\)

   • Integrate as normal, treating the inside as x (the integral of cos is sin): \(\sin(6x+2)\).
   • Divide by 6, the coefficient of x: \(1\div6=\frac{1}{6}\).
   • \(\int \cos(6x+2)\,dx=\frac{1}{6}\sin(6x+2)+c\)

2. (b) \(\int 8\cos(4x-5)\,dx\)

   • Integrate as normal, treating the inside as x (the integral of cos is sin): \(8\sin(4x-5)\).
   • Divide by 4, the coefficient of x: \(8\div4=2\).
   • \(\int 8\cos(4x-5)\,dx=2\sin(4x-5)+c\)

Find each integral.

1. (a) \(\int \sin(5x-1)\,dx\)

   • Integrate as normal, treating the inside as x (the integral of sin is minus cos): \(-\cos(5x-1)\).
   • Divide by 5, the coefficient of x: \(-1\div5=-\frac{1}{5}\).
   • \(\int \sin(5x-1)\,dx=-\frac{1}{5}\cos(5x-1)+c\)

2. (b) \(\int 9\sin(2-3x)\,dx\)

   • Integrate as normal, treating the inside as x (the integral of sin is minus cos): \(-9\cos(2-3x)\).
   • Divide by -3, the coefficient of x: \(-9\div(-3)=3\).
   • \(\int 9\sin(2-3x)\,dx=3\cos(2-3x)+c\)

Find each integral.

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

   • Integrate as normal, treating the inside as x (the integral of e to a power is itself): \(6e^{3x-2}\).
   • Divide by 3, the coefficient of x: \(6\div3=2\).
   • \(\int 6e^{3x-2}\,dx=2e^{3x-2}+c\)

2. (b) \(\int 12e^{4-3x}\,dx\)

   • Integrate as normal, treating the inside as x (the integral of e to a power is itself): \(12e^{4-3x}\).
   • Divide by -3, the coefficient of x: \(12\div(-3)=-4\).
   • \(\int 12e^{4-3x}\,dx=-4e^{4-3x}+c\)

Find each integral.

1. (a) \(\int \frac{10}{5x-1}\,dx\)

   • Integrate as normal, treating the inside as x (the integral of 1/u is ln|u|): \(10\ln|5x-1|\).
   • Divide by 5, the coefficient of x: \(10\div5=2\).
   • \(\int \frac{10}{5x-1}\,dx=2\ln|5x-1|+c\)

2. (b) \(\int \frac{8}{3-4x}\,dx\)

   • Integrate as normal, treating the inside as x (the integral of 1/u is ln|u|): \(8\ln|3-4x|\).
   • Divide by -4, the coefficient of x: \(8\div(-4)=-2\).
   • \(\int \frac{8}{3-4x}\,dx=-2\ln|3-4x|+c\)

Find \(\int 10x(x^2+4)^{3}\,dx\).

• Spot the pair: \(g(x)=x^2+4\) and \(g'(x)=2x\).
• The integrand is \(5g'(x)[g(x)]^{3}\), so the integral is \(5\times\frac{(x^2+4)^{4}}{4}\).
• \(\int 10x(x^2+4)^{3}\,dx=\frac{5}{4}(x^2+4)^{4}+c\)
• Check: differentiate the answer and you get the integrand back.

Find each integral.

1. (a) \(\int 6\cos x\sin^{2}x\,dx\)

   • \(\cos x\) is the derivative of \(\sin x\): take \(g(x)=\sin x\).
   • The integral is \(6\times\frac{\sin^{3}x}{3}\).
   • \(\int 6\cos x\sin^{2}x\,dx=2\sin^{3}x+c\)
   • Check: differentiate the answer and you get the integrand back.

2. (b) \(\int \sin x\cos^{5}x\,dx\)

   • \(\sin x\) is minus the derivative of \(\cos x\): take \(g(x)=\cos x\), \(g'(x)=-\sin x\).
   • Allow for the minus sign: the integral is \(-\frac{\cos^{6}x}{6}\).
   • \(\int \sin x\cos^{5}x\,dx=-\frac{1}{6}\cos^{6}x+c\)
   • Check: differentiate the answer and you get the integrand back.

Find each integral.

1. (a) \(\int \frac{6x}{x^2+2}\,dx\)

   • Spot the pair: \(g(x)=x^2+2\) and \(g'(x)=2x\).
   • The top is \(3g'(x)\), so the integral is \(3\ln|g(x)|\).
   • \(\int \frac{6x}{x^2+2}\,dx=3\ln|x^2+2|+c\)
   • Check: differentiate the answer and you get the integrand back.

2. (b) \(\int \frac{4x^{3}}{x^4+9}\,dx\)

   • Spot the pair: \(g(x)=x^4+9\) and \(g'(x)=4x^{3}\).
   • The top is exactly \(g'(x)\), so the integral is \(\ln|g(x)|\).
   • \(\int \frac{4x^{3}}{x^4+9}\,dx=\ln|x^4+9|+c\)
   • Check: differentiate the answer and you get the integrand back.

Find \(\int 9x^{2}e^{x^3+2}\,dx\).

• Spot the pair: \(g(x)=x^3+2\) and \(g'(x)=3x^{2}\).
• The integrand is \(3g'(x)e^{g(x)}\), so the integral is \(3e^{g(x)}\).
• \(\int 9x^{2}e^{x^3+2}\,dx=3e^{x^3+2}+c\)
• Check: differentiate the answer and you get the integrand back.

Find \(\int \frac{15}{\cos^2(5x+1)}\,dx\).

• \(\int\frac{1}{\cos^2 u}\,du=\tan u\)
• Divide by 5, the coefficient of x: \(15\div 5=3\)
• \(\int\frac{15}{\cos^2(5x+1)}\,dx=3\tan(5x+1)+c\)
• Check: differentiate and you get the integrand back.

Oil leaks from a tank at the rate \(\frac{dV}{dt}=30e^{-0.3t}\) litres per minute, where \(V\) litres is the amount leaked and \(t\) is the time in minutes. When \(t=0\), \(V=0\).

1. (a) Find \(V\) in terms of \(t\).

   • \(V=\int 30e^{-0.3t}\,dt=-100e^{-0.3t}+c\): divide by \(-0.3\), the coefficient of t.
   • When \(t=0\), \(V=0\): \(0=-100+c\), so \(c=100\).

2. (b) Find \(V\) when \(t=6\), to 3 significant figures.

   • \(V=-100e^{-0.3\times 6}+100=83.5\) litres

The number of visitors to a website, \(V\), is growing at the rate \(\frac{dV}{dt}=\frac{450}{3t+2}\) per day, where \(t\) is the time in days. When \(t=0\), \(V=500\).

1. (a) Find \(V\) in terms of \(t\).

   • \(V=\int\frac{450}{3t+2}\,dt=150\ln(3t+2)+c\): divide by 3, the coefficient of t.
   • When \(t=0\), \(V=500\): \(500=150\ln 2+c\), so \(c=500-150\ln 2\).

2. (b) Find \(V\) when \(t=8\), to 3 significant figures.

   • \(V=150\ln(3\times 8+2)+500-150\ln 2=885\) visitors

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

• Spot the pair: \(g(x)=x^3+1\) and \(g'(x)=3x^{2}\).
• The integrand is \(2g'(x)[g(x)]^{\frac{1}{2}}\), so the integral is \(2\times\frac{(x^3+1)^{\frac{3}{2}}}{\frac{3}{2}}\).
• \(\int 6x^2\sqrt{x^3+1}\,dx=\frac{4}{3}(x^3+1)^{\frac{3}{2}}+c\)
• Check: differentiate the answer and you get the integrand back.

Given that \(f'(x)=\frac{6\sin x}{3-\cos x}\) and \(f(\pi)=5\), find \(f(x)\), with the constant to 3 significant figures.

• The top is a multiple of the derivative of the bottom: \(g=3-\cos x\), \(g'=\sin x\).
• \(f(x)=6\ln(3-\cos x)+c\). The bottom is always positive, so no modulus is needed.
• Put in the point: \(5=6\ln(4)+c\), so \(c=5-6\ln 4\).
• \(c=-3.32\), so \(f(x)=6\ln(3-\cos x)-3.32\)