Examples for Section 7.2
Integration by Parts

p. 567

5.

Set u  =  ln(x), dv  =  x dx, du  =  1/x, v  =  (1/2) (x^2). Then
∫x ln(x) dx = (1/2) (x^2) ln(x)  -  ∫((1/2) (x^2))((1/x)) dx = (1/2) (x^2) ln(x)  - (1/4) (x^4) +  C

∫x Log[x] \[DifferentialD]x

15.

Use tabular integration (see p. 567): The derivatives of (x^5)are 5  (x^4), 20  (x^3), 60  (x^2), 120  x, 120. The integrals of (e^x)are all (e^x). Pair the derivative with the integrals ("diagonally") with alternating signs to get: \(TraditionalForm
      \`∫\(x\^5\) \(e\^x\) dx\  = \ 
        \(x\^5\) e\^x - \ 5 \( x\^4\) e\^x + \ 20 \( x\^3\) e\^x - 
          60 \( x\^2\) e\^x + \ 120  x\ e\^x - \ 120  e\^x\)

∫(x^5) (E^x) \[DifferentialD]x

21.

Let u  =  (e^θ), dv  =  sin(θ), \(TraditionalForm
      \`du\  = \ \(e\^θ\) dθ\), v  =  -cos(θ). Then \(TraditionalForm
      \`∫\(e\^θ\) \(sin(θ)\) 
            dθ\  = \ 
        \(-e\^θ\) \(cos(θ)\)\  + \ 
          ∫\(e\^θ\) \(cos(θ)\) 
              dθ\). Repeat with u  =  (e^θ), \(TraditionalForm
      \`dv\  = \ 
        \(cos(θ)\) dθ\), to get\(TraditionalForm
      \`∫\(e\^θ\) \(sin(θ)\) 
            dθ\  = \ 
        \(\(-e\^θ\) \(cos(θ)\)\  + \ 
            ∫\(e\^θ\) \(cos(θ)\) 
                dθ\  = 
          \(-e\^θ\) \(cos(θ)\)\ \  + \ 
            \(e\^θ\) \(sin(θ)\)\ \  - 
            ∫\(e\^θ\) \(sin(θ)\) 
                dθ\)\) so \(TraditionalForm
      \`\(2 \(∫\(e\^θ\) \(sin(θ)\) 
              dθ\)\  = 
        \(-e\^θ\) \(cos(θ)\)\ \  + \ 
          \(e\^θ\) \(sin(θ)\)\  + \ C\_0\ \ \)\)and \(TraditionalForm
      \`∫\(e\^θ\) \(sin(θ)\) 
            dθ\  = 
        \(1\/2\) \(\(e\^θ\)(sin(θ) - cos(θ))\)\  + \ C
          \).

∫(E^θ) Sin[θ] \[DifferentialD]θ

29.

Substitute u = ln(x), du = (1/x) dx, \(TraditionalForm
      \`dx = 
        \(x\ du = \(e\^u\) du\)\), to get \(TraditionalForm
      \`∫\(sin(ln(x))\) dx = 
        \(∫\(sin(u)\) \(e\^u\) du = 
          \(1\/2\) \(\(e\^u\)(sin(u) - cos(u))\)\  + \ C\)\) (see problem 21). Substituting back u  =  ln(x) gives \(TraditionalForm
      \`\(\(∫\(sin(ln(x))\) dx = 
                  \(1\/2\) \(\(e\^\(ln(x)\)\)(sin(ln(x)) - cos(ln(x)))\))
                \)\  + \ C\  = \(1\/2\) \(x(sin(ln(x)) - cos(ln(x)))\))\)\  + 
        \ C\)

∫Sin[Log[x]] \[DifferentialD]x

42.

The average value of a function f(x) on an interval [a, b] is given by \(TraditionalForm
      \`avg(f)\  = \ 
        \(1\/\(b - a\)\) 
          \(∫\_a\%b\( f(x)\) dx\)\). So the average value of \(TraditionalForm
      \`f(t)\  = \ 4 \(\( e\^\(-t\)\)(sin(t) - cos(t))\)\) on [0, 2  π] is \(TraditionalForm
      \`\(1\/\(2  π\)\) 
        \(∫4 \(\( e\^\(-t\)\)(sin(t) - cos(t))\) 
            dt\)\). Using integration by parts, we get avg(f)  = - (2/π)(e^(-t))sin(t) \(|\_0\%\(2  π\)\)= 0.

(1/(2  π)) 
      ∫ 
        4  (E^(-t)) (Sin[t] - Cos[t]) \[DifferentialD]t

(1/(2  π)) 
      (∫_0^(2  π)) 
        4  (E^(-t)) (Sin[t] - Cos[t]) \[DifferentialD]t

\(Plot[{\ 4 \( E\^\(-t\)\) \((Sin[t] - Cos[t])\), \ 0}, {t, 0, 2  π}, 
      
	PlotStyle -> {RGBColor[0, 0, 1], \ RGBColor[1, 0, 0]}, \ 
      PlotRange -> {\(-4\), 2}]\)