Integration by substitution Problem set B Worked solutions

Use the given substitution to find each integral.

1. (a) \(\int \cos 7x\,dx,\quad u=7x\)

   • \(u=7x\), so \(du=7\,dx\)
   • \(\int \tfrac17\cos u\,du=\tfrac17\sin u\)
   • \(\tfrac17\sin 7x+c\)

2. (b) \(\int \frac{1}{4-5x}\,dx,\quad u=4-5x\)

   • \(u=4-5x\), so \(du=-5\,dx\)
   • \(\int -\tfrac15\cdot\frac1u\,du=-\tfrac15\ln|u|\)
   • \(-\tfrac15\ln|4-5x|+c\)

3. (c) \(\int e^{3+2x}\,dx,\quad u=3+2x\)

   • \(u=3+2x\), so \(du=2\,dx\)
   • \(\int \tfrac12e^u\,du=\tfrac12e^u\)
   • \(\tfrac12e^{3+2x}+c\)

Use the given substitution to find each integral.

1. (a) \(\int \frac{e^x}{6+e^x}\,dx,\quad u=6+e^x\)

   • \(u=6+e^x\), so \(du=e^x\,dx\)
   • \(\int \frac1u\,du=\ln u\)
   • \(\ln(6+e^x)+c\)

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

   • \(u=2+x^4\), so \(du=4x^3\,dx\)
   • \(\int \tfrac14\cdot\frac1u\,du=\tfrac14\ln|u|\)
   • \(\tfrac14\ln|2+x^4|+c\)

Use the given substitution to find each integral.

1. (a) \(\int x\sqrt{x^2+11}\,dx,\quad u=x^2+11\)

   • \(u=x^2+11\), so \(du=2x\,dx\)
   • \(\int \tfrac12\sqrt u\,du=\tfrac13u^{3/2}\)
   • \(\tfrac13(x^2+11)^{3/2}+c\)

2. (b) \(\int \cos x\sqrt{1+\sin x}\,dx,\quad u=1+\sin x\)

   • \(u=1+\sin x\), so \(du=\cos x\,dx\)
   • \(\int \sqrt u\,du=\tfrac23u^{3/2}\)
   • \(\tfrac23(1+\sin x)^{3/2}+c\)

Use the given substitution to find each integral.

1. (a) \(\int \frac{e^{-\sqrt x}}{\sqrt x}\,dx,\quad u=\sqrt x\)

   • \(u=\sqrt x\), so \(du=\frac{1}{2\sqrt x}\,dx\)
   • \(\int 2e^{-u}\,du=-2e^{-u}\)
   • \(-2e^{-\sqrt x}+c\)

2. (b) \(\int \frac{1}{x(1+\ln x)}\,dx,\quad u=1+\ln x\)

   • \(u=1+\ln x\), so \(du=\frac1x\,dx\)
   • \(\int \frac1u\,du=\ln|u|\)
   • \(\ln|1+\ln x|+c\)

Choose a substitution to find each integral.

1. (a) \(\int \sin 8x\,dx\)

   • \(u=8x\), so \(du=8\,dx\)
   • \(\int \tfrac18\sin u\,du=-\tfrac18\cos u\)
   • \(-\tfrac18\cos 8x+c\)

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

   • \(u=-4x\), so \(du=-4\,dx\)
   • \(\int -\tfrac14e^u\,du=-\tfrac14e^u\)
   • \(-\tfrac14e^{-4x}+c\)

3. (c) \(\int \frac{1}{2+5x}\,dx\)

   • \(u=2+5x\), so \(du=5\,dx\)
   • \(\int \tfrac15\cdot\frac1u\,du=\tfrac15\ln|u|\)
   • \(\tfrac15\ln|2+5x|+c\)

4. (d) \(\int \frac{1}{9-x}\,dx\)

   • \(u=9-x\), so \(du=-1\,dx\)
   • \(\int -\frac1u\,du=-\ln|u|\)
   • \(-\ln|9-x|+c\)

Choose a substitution to find each integral.

1. (a) \(\int x^7e^{x^8}\,dx\)

   • \(u=x^8\), so \(du=8x^7\,dx\)
   • \(\int \tfrac18e^u\,du=\tfrac18e^u\)
   • \(\tfrac18e^{x^8}+c\)

2. (b) \(\int x^3\cos(x^4)\,dx\)

   • \(u=x^4\), so \(du=4x^3\,dx\)
   • \(\int \tfrac14\cos u\,du=\tfrac14\sin u\)
   • \(\tfrac14\sin(x^4)+c\)

3. (c) \(\int \frac{x}{5+4x^2}\,dx\)

   • \(u=5+4x^2\), so \(du=8x\,dx\)
   • \(\int \tfrac18\cdot\frac1u\,du=\tfrac18\ln u\)
   • \(\tfrac18\ln(5+4x^2)+c\)

4. (d) \(\int \frac{\sin x}{3+\cos x}\,dx\)

   • \(u=3+\cos x\), so \(du=-\sin x\,dx\)
   • \(\int -\frac1u\,du=-\ln u\)
   • \(-\ln(3+\cos x)+c\)

Choose a substitution to find each integral. In the second part, write tan as sin over cos.

1. (a) \(\int \cos^3x\sin x\,dx\)

   • \(u=\cos x\), so \(du=-\sin x\,dx\)
   • \(\int -u^3\,du=-\tfrac14u^4\)
   • \(-\tfrac14\cos^4x+c\)

2. (b) \(\int \tan 3x\,dx\)

   • Write \(\tan\) as \(\frac{\sin}{\cos}\).
   • \(u=\cos 3x\), so \(du=-3\sin 3x\,dx\)
   • \(\int -\tfrac13\cdot\frac1u\,du=-\tfrac13\ln|u|\)
   • \(-\tfrac13\ln|\cos 3x|+c\)

Oil flows into a tank at a rate of \(R=t\sqrt{t^2+16}\) litres per minute, where t is the time in minutes. Find the volume of oil that flows in during the first 3 minutes.

• \(u=t^2+16\), \(du=2t\,dt\); limits \(t=0\Rightarrow u=16\), \(t=3\Rightarrow u=25\)
• \(\int_{16}^{25}\frac12\sqrt u\,du=\frac13\left[u^{3/2}\right]_{16}^{25}\)
• \(\frac13(125-64)=\frac{61}{3}=20.3\) litres

The concentration C of a drug in the blood, in mg per litre, at time t hours satisfies \(\frac{dC}{dt}=-8e^{-0.2t}\). At \(t=0\), \(C=30\). Find C when \(t=6\).

• \(C=\int-8e^{-0.2t}dt\). Let \(u=-0.2t\), \(du=-0.2\,dt\)
• \(C=\frac{-8}{-0.2}e^{-0.2t}+c=40e^{-0.2t}+c\)
• \(30=40+c\), so \(c=-10\)
• \(C=40e^{-1.2}-10=2.05\) mg per litre

A garden bed is the region under \(y=\frac{e^x}{5+e^x}\) from \(x=0\) to \(x=2\) (units in metres). Find its area, then check with your calculator.

• \(u=5+e^x\), \(du=e^x\,dx\); limits \(u=6\) to \(u=5+e^2\)
• \(\int_6^{5+e^2}\frac1u\,du=\left[\ln u\right]_6^{5+e^2}\)
• \(\ln(5+e^2)-\ln6=0.725\) m²

Use \(u=x+5\) to find \(\int\sqrt{x^3+5x^2}\,dx\), for \(x>0\).

• \(\sqrt{x^3+5x^2}=x\sqrt{x+5}\) for \(x>0\)
• \(x=u-5\), \(dx=du\): \(\int(u-5)\sqrt u\,du=\int(u^{3/2}-5u^{1/2})du\)
• \(\frac25u^{5/2}-\frac{10}{3}u^{3/2}+c=\frac25(x+5)^{5/2}-\frac{10}{3}(x+5)^{3/2}+c\)

Use \(x=\sin u\) to evaluate \(\int_0^{\sqrt2/2}\frac{1}{\sqrt{1-x^2}}\,dx\).

• \(dx=\cos u\,du\) and \(\sqrt{1-\sin^2u}=\cos u\), so the integrand becomes \(du\)
• Limits: \(x=0\Rightarrow u=0\); \(x=\frac{\sqrt2}{2}\Rightarrow u=\frac{\pi}{4}\)
• \(\left[u\right]_0^{\pi/4}=\frac{\pi}{4}=0.785\)