Examples for Section 7.4
Trigonometric Substitutions

p. 582

7.

Use a trig substitution t  =  5  sin(θ), dt  =  5  cos(θ), (Sqrt[25 - (t^2)]) =  5  cos(θ). Then \(TraditionalForm
      \`∫\(\@\(25\  - \ t\^2\)\) dt\  = \ 
        \(∫5 \( cos(θ)\) 5 \( cos(θ)\) 
              dθ\  = \ 
          \(25\/2\) \(∫1\)\  + \ 
            \(cos(2  θ)\) dθ\)\). Here we used the identity \(TraditionalForm
      \`\(cos\^2\)(θ)\  = \ 
        \(1\/2\) \((1\  + \ cos(2  θ))\)\). Integration yields \(TraditionalForm
      \`\(25\/2\) \((θ\  + \ \(1\/2\) \(sin(2  θ)\))\) + C\). To put this back in terms of θ we rewrite \(TraditionalForm
      \`sin(2  θ)\  = \ 2 \( sin(θ)\) \(cos(θ)\)\): \(TraditionalForm
      \`\(25\/2\) \((θ\  + \ \(1\/2\) \(sin(2  θ)\))\)\  = \ 
        \(\(25\/2\) 
            \((θ + \(sin(θ)\) \(cos(θ)\))\)\ \  = \ 
          \(\(25\/2\) 
              \((\(sin\^\(-1\)\)(t\/5)\  + \ \(t\/5\) \@\(25 - t\^2\)\/5)
                \)\  = \ 
            \(25\/2\) \(\(sin\^\(-1\)\)(θ)\)\  + \ 
              \(t\/2\) \@\(25 - t\^2\)\)\)\)

∫(Sqrt[25  -  (t^2)]) \[DifferentialD]t

15.

Use a trig substitution x  =  2  tan(θ), \(TraditionalForm
      \`dx\  = \ 
        2 \(\( sec\^2\)(θ)\) dθ\), (Sqrt[(x^2) + 4]) =  2  sec(θ). Then \(TraditionalForm
      \`∫\(x\^3\/\@\(x\^2 + 4\)\) dx = \ 
        \(∫\(\(8 \(\( tan\^3\)(θ)\)\)\/\(2 
                    \( sec(θ)\)\)\) 2 \(\( sec\^2\)(θ)\) 
              dθ\  = \ 
          \(8 \(∫\(\(tan\^3\)(θ)\) \(sec(θ)\) 
                  dθ\)\  = \ 
            8 \(∫\((\(sec\^2\)(θ) - 1)\) \(sec(θ)\) 
                  \(tan(θ)\) dθ\)\)\)\)
Substituting u  =  sec(θ), \(TraditionalForm
      \`du\  = \ 
        \(sec(θ)\) \(tan(θ)\) dθ\): \(TraditionalForm
      \`8 \(∫\((u\^2 - 1)\) du\) = 
        \(\(8\/3\) u\^3 - 8  u + C\  = \ 
          \(\(8\/3\) \(\(sec\^3\)(θ)\)\  - \ 8 \( sec(θ)\) + C
              \  = \ \(\(1\/3\) \((x\^2 + 4)\)\^\(3/2\) - \ 
                4 \@\( x\^2 + 4\) + \ C\  = \ 
              \(1\/3\) \((x\^2 - 8)\) \@\(x\^2 + 4\) + C\)\)\)\)

∫((x^3)/(Sqrt[(x^2) + 4])) \[DifferentialD]x  // Simplify

29.

Substitute u  =  (e^t), \(TraditionalForm
      \`du\  = \ \(e\^t\) dt\), to get \(TraditionalForm
      \`∫\(e\^t\/\@\(e\^\(2  t\) + 9\)\) dt\  = \ 
        ∫\(1\/\@\(u\^2 + 9\)\) du\).  Now use a trig substitution, u  =   3  tan(θ), \(TraditionalForm
      \`du\  = \ 
        3 \(\( sec\^2\)(θ)\) dθ\), (Sqrt[(u^2) + 9]) =  3  sec(θ): \(TraditionalForm
      \`∫\(1\/\@\(u\^2 + 9\)\) du\  = \ \ 
        \(∫\(\(3 \(\( sec\^2\)(θ)\)\)\/\(3 
                    \( sec(θ)\)\)\) dθ\  = \ 
          \(∫\(sec(θ)\) dθ\  = \ 
            \(ln | sec(θ) + tan(θ) | \(+C\)\  = \ 
              \(ln | \@\(e\^\(2  t\) + 9\)\/3 + e\^t\/3 | \(+C\)\  = \ \ 
                ln | \@\(e\^\(2  t\) + 9\) + e\^t | \(+C\_1\)\)\)\)\)\)
where (Subscript[C, 1]) =  C  -  ln(3). The definite integral is then \(TraditionalForm
      \`∫\_0\%\(ln(4)\)\(e\^t\/\@\(e\^\(2  t\) + 9\)\) 
            dt\  = 
        \(\((ln | \@\(e\^\(2  t\) + 9\) + e\^t | )\)\_0\%\(ln(4)\) = 
          \(ln | \@\(4\^2 + 9\) + 4 | \(-ln\) | \ \@\(1 + 9\) + 1 | \  = \ 
            ln(9)\  - \ ln(\@10 + 1)\)\)\)

(∫_0^(Log[4]))((E^t)/(Sqrt[(E^(2  t)) + 9])) 
        \[DifferentialD]t

N[-ArcSinh[(1/3)] + ArcSinh[(4/3)]]

N[Log[9]  -  Log[(Sqrt[10]) + 1]]