Examles for Section 7.1
Basic Integration Formulas

p. 560

3.

Substitute u  =  sin(v),  
      du  =  cos(v) dv to get ∫3 (Sqrt[ u]) du.

∫3 (Sqrt[Sin])[v] Cos[v] \[DifferentialD]v

7.

Substitute u  =  (Sqrt[x]) + 1, \(TraditionalForm
      \`du\  = \ \(1\/\(2 \@ x\)\) dx\)to get 2 ∫(1/u) du.

∫(1/((Sqrt[x]) ((Sqrt[x]) +  1))) \[DifferentialD]x

33.

Multiply numerator and denominator by (e^x)to get \(TraditionalForm
      \`∫\(e\^x\/\(e\^\(2  x\) + \ 1\)\) dx\) and then substitute u  = (e^x), \(TraditionalForm
      \`du\  = \ \(e\^x\) dx\) to get \(TraditionalForm
      \`∫du\/\(u\^2 + \ 1\) = \ 
        \(tan\^\(-1\)\)(u)\  + \ C\).

∫(1/((E^x) +  (E^(-x)))) \[DifferentialD]x

Simplify[%]

37.

Complete the square to get \(TraditionalForm
      \`∫\_1\%2
          \( 8 dx\)\/\(\((x - 1)\)\^2 + 1\)\)and then substitute u  =  x - 1.

∫((8 \[DifferentialD]x)/((((x - 1))^2) + 1))

39.

Complete the square to get \(TraditionalForm
      \`∫dt\/\@\(1\  - \ \((t - 2)\)\^2\)\) then substitute u  =  t - 2.

∫\[DifferentialD](t/(Sqrt[1  -  (((t - 1))^2)]))

49.

Divide (x^2) - 1 into 2  (x^3)to get \(TraditionalForm
      \`∫\(\(2  x\^3\)\/\(x\^2 - 1\)\) dx\  = \ 
        ∫2  x dx\  + \ 
          ∫\(\(2  x\)\/\(x\^2 - 1\)\) dx\) then use a substitution u  =  (x^2)  -  1in the second integral.

∫2  x \[DifferentialD]x  +  
      ∫((2  x)/((x^2) - 1)) \[DifferentialD]x

53.

Separate the integrand into two fractions, \(TraditionalForm
      \`\(1 - x\)\/\@\(x\^2 - 1\)\  = \ 
        1\/\@\(x\^2 - \ 1\) - \ x\/\@\(x\^2 - \ 1\)\), then integrate the first term as \(TraditionalForm
      \`\(cosh\^\(-1\)\)(x)\  = \ ln(x\  + \ \@\(x\^2 - \ 1\))\) and the second term with a substitution u  =  (x^2) -  1.

∫((1 - x)/(Sqrt[(x^2) - 1])) \[DifferentialD]x

57.

Multiply the numerator and denominator by 1  -  sin(x) to get \(TraditionalForm
      \`\(1\/\(1 + sin(x)\) = 
        \(\(\ 1\  - \ sin(x)\)\/\(1\  - \ \(sin\^2\)(x)\) = 
          \(\(\ 1\  - \ sin(x)\)\/\(\(cos\^2\)(x)\)\  = \ 
            \(sec\^2\)(x)\  - \(sec(x)\) \(tan(x)\)\)\)\ \)\). The integral of (sec^2)(x)is tan(x) and the integral of sec(x) tan(x) is sec(x) so \(TraditionalForm
      \`∫\(1\/\(1 + sin(x)\)\) dx\  = \ 
        tan(x)\  - \ sec(x)\  + \ C\).

∫(Sec[x]^2  -  Sec[x] Tan[x]) \[DifferentialD]x

∫(1/(1 + Sin[x])) \[DifferentialD]x

These expressions all represent the same function, tan(x)  -  sec(x).

83.

\(TraditionalForm
      \`∫\(\(cos\^3\)(x)\)\ dx\  = \ 
        \(∫\(cos(x)\) \((1 - \(sin\^2\)(x))\) dx\  = 
          \ ∫\(cos(x)\) dx\  - \ 
            ∫\ 
              \(\(sin\^2\)(x)\) \(cos(x)\) dx\)\) The second integral can be interated with a substitution u  =  sin(x), \(TraditionalForm
      \`du\  = \ \(cos(x)\) dx\).

∫Cos[x]^3 \[DifferentialD]x

TrigExpand[%]

%  /.  Cos[x]^2  -> 1  -  Sin[x]^2 // Simplify

∫Cos[x] \[DifferentialD]x  -  
      ∫Sin[x]^2 Cos[x] \[DifferentialD]x