Integration by substitution Problem set A One question at a time

Use the given substitution to find each integral.

1. (a) \(\int \sin 9x\,dx,\quad u=9x\)

   • \(u=9x\), so \(du=9\,dx\)
   • \(\int \tfrac19\sin u\,du=-\tfrac19\cos u\)
   • \(-\tfrac19\cos 9x+c\)

2. (b) \(\int \frac{1}{7-3x}\,dx,\quad u=7-3x\)

   • \(u=7-3x\), so \(du=-3\,dx\)
   • \(\int -\tfrac13\cdot\frac1u\,du=-\tfrac13\ln|u|\)
   • \(-\tfrac13\ln|7-3x|+c\)

3. (c) \(\int e^{2+5x}\,dx,\quad u=2+5x\)

   • \(u=2+5x\), so \(du=5\,dx\)
   • \(\int \tfrac15e^u\,du=\tfrac15e^u\)
   • \(\tfrac15e^{2+5x}+c\)

Use the given substitution to find each integral.

1. (a) \(\int \frac{e^x}{4+e^x}\,dx,\quad u=4+e^x\)

   • \(u=4+e^x\), so \(du=e^x\,dx\)
   • \(\int \frac1u\,du=\ln u\)
   • \(\ln(4+e^x)+c\)

2. (b) \(\int \frac{x^2}{5+x^3}\,dx,\quad u=5+x^3\)

   • \(u=5+x^3\), so \(du=3x^2\,dx\)
   • \(\int \tfrac13\cdot\frac1u\,du=\tfrac13\ln|u|\)
   • \(\tfrac13\ln|5+x^3|+c\)

Use the given substitution to find each integral.

1. (a) \(\int x\sqrt{x^2+7}\,dx,\quad u=x^2+7\)

   • \(u=x^2+7\), so \(du=2x\,dx\)
   • \(\int \tfrac12\sqrt u\,du=\tfrac13u^{3/2}\)
   • \(\tfrac13(x^2+7)^{3/2}+c\)

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

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

Use the given substitution to find each integral.

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

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

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

   • \(u=\ln x\), so \(du=\frac1x\,dx\)
   • \(\int \sin u\,du=-\cos u\)
   • \(-\cos(\ln x)+c\)

Choose a substitution to find each integral.

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

   • \(u=6x\), so \(du=6\,dx\)
   • \(\int \tfrac16\cos u\,du=\tfrac16\sin u\)
   • \(\tfrac16\sin 6x+c\)

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

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

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

   • \(u=5+2x\), so \(du=2\,dx\)
   • \(\int \tfrac12\cdot\frac1u\,du=\tfrac12\ln|u|\)
   • \(\tfrac12\ln|5+2x|+c\)

4. (d) \(\int \frac{1}{6-x}\,dx\)

   • \(u=6-x\), so \(du=-1\,dx\)
   • \(\int -\frac1u\,du=-\ln|u|\)
   • \(-\ln|6-x|+c\)

Choose a substitution to find each integral.

1. (a) \(\int x^3e^{x^4}\,dx\)

   • \(u=x^4\), so \(du=4x^3\,dx\)
   • \(\int \tfrac14e^u\,du=\tfrac14e^u\)
   • \(\tfrac14e^{x^4}+c\)

2. (b) \(\int x^5\sin(x^6)\,dx\)

   • \(u=x^6\), so \(du=6x^5\,dx\)
   • \(\int \tfrac16\sin u\,du=-\tfrac16\cos u\)
   • \(-\tfrac16\cos(x^6)+c\)

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

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

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

   • \(u=4+\sin x\), so \(du=\cos x\,dx\)
   • \(\int \frac1u\,du=\ln u\)
   • \(\ln(4+\sin x)+c\)

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

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

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

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

   • Write \(\tan\) as \(\frac{\sin}{\cos}\).
   • \(u=\cos 2x\), so \(du=-2\sin 2x\,dx\)
   • \(\int -\tfrac12\cdot\frac1u\,du=-\tfrac12\ln|u|\)
   • \(-\tfrac12\ln|\cos 2x|+c\)

A particle moves in a straight line with velocity \(v=t\sqrt{t^2+9}\) m s\(^{-1}\), where t is the time in seconds. Find the distance travelled in the first 4 seconds.

• \(u=t^2+9\), \(du=2t\,dt\); limits \(t=0\Rightarrow u=9\), \(t=4\Rightarrow u=25\)
• \(\int_9^{25}\frac12\sqrt u\,du=\frac13\left[u^{3/2}\right]_9^{25}\)
• \(\frac13(125-27)=\frac{98}{3}=32.7\) m

A hot drink cools. Its temperature T, in °C, at time t minutes satisfies \(\frac{dT}{dt}=-12e^{-0.3t}\). At \(t=0\), \(T=90\). Find T when \(t=5\).

• \(T=\int-12e^{-0.3t}dt\). Let \(u=-0.3t\), \(du=-0.3\,dt\)
• \(T=\frac{-12}{-0.3}e^{-0.3t}+c=40e^{-0.3t}+c\)
• \(90=40+c\), so \(c=50\)
• \(T=40e^{-1.5}+50=58.9\) °C

The cross-section of a ramp is the region under \(y=\frac{x}{4+x^2}\) from \(x=0\) to \(x=2\) (units in metres). Find its area, then check with your calculator.

• \(u=4+x^2\), \(du=2x\,dx\); limits \(u=4\) to \(u=8\)
• \(\int_4^8\frac12\cdot\frac1u\,du=\frac12\left[\ln u\right]_4^8\)
• \(\frac12(\ln8-\ln4)=\frac12\ln2=0.347\) m²

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

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

Use \(x=\sin u\) to evaluate \(\int_0^{\sqrt3/2}\frac{1}{\sqrt{1-x^2}}\,dx\).

• \(dx=\cos u\,du\) and \(\sqrt{1-\sin^2u}=\cos u\), so the integrand becomes \(du\)
• Limits: \(x=0\Rightarrow u=0\); \(x=\frac{\sqrt3}{2}\Rightarrow u=\frac{\pi}{3}\)
• \(\left[u\right]_0^{\pi/3}=\frac{\pi}{3}=1.05\)