p. 591

2.

∫\[DifferentialD](x/(x (Sqrt[ x + 4])))

8.

∫\[DifferentialD](x/((x^2) (Sqrt[4  x - 9])))

16.

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

20.

∫\[DifferentialD](θ/(4  +  
          5  Sin[2  θ]))

24.

∫x ArcSin[x] \[DifferentialD]x

34.

∫Sin[2  x] Cos[3  x] \[DifferentialD]x

54.

∫Sin[(θ/2)]^5 \[DifferentialD]θ

62.

∫Tan[(x/2)]^4 \[DifferentialD]x

72.

∫Log[x]^3 \[DifferentialD]x

109.

a)

∫x Log[x] \[DifferentialD]x

b)

∫(x^2) Log[x] \[DifferentialD]x

c)

∫(x^3) Log[x] \[DifferentialD]x

d) The pattern looks like \(TraditionalForm
      \`∫\(x\^n\) \(ln(x)\) dx = \ 
        \(-\(x\^\(n + 1\)\/\((n + 1)\)\^2\)\) + 
          \(1\/\(n + 1\)\) \(ln(x)\)\), so for n = 4the answer should be - ((x^5)/25) + (1/5) ln(x).

∫(x^4) Log[x] \[DifferentialD]x

e) Mathematica knows the general formula:

∫(x^n) Log[x] \[DifferentialD]x

110.

a)

∫(( Log[x])/(x^2)) \[DifferentialD]x

b)

∫(( Log[x])/(x^3)) \[DifferentialD]x

c)

∫(( Log[x])/(x^4)) \[DifferentialD]x

d) The pattern looks like \(TraditionalForm
      \`∫\(\(ln(x)\)\/x\^n\) dx = \ 
        \(-\(1\/\(\(\((n - 1)\)\^2\) x\^\(n - 1\)\)\)\) - 
          \(1\/\(n - 1\)\) \(ln(x)\)\), so for n = 5the answer should be - (1/(16  (x^4))) - (1/4) ln(x).

∫(( Log[x])/(x^5)) \[DifferentialD]x

e) Mathematica knows the general formula:

∫ Log[x]/(x^n) \[DifferentialD]x

111.

a) Mathematica cannot compute:

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
        \[DifferentialD]x\)

b)

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 1\)

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 2\)

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 3\)

Simplify[%]

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 4\)

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 5\)

Simplify[%]

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 6\)

\(∫\_0\%\(π/2\)\(Sin[x]\^n\/\(Sin[x]\^n + \ Cos[x]\^n\)\) 
          \[DifferentialD]x\  /. \ n -> 7\)

Mathematica gives up at n = 7.

c) Substituting x  =  π/2  -  u, \(TraditionalForm
      \`dx\  = \ \(-\ du\)\), and using the identities sin(π/2 - u)  =  cos(u) and cos(π/2 - u)  =  sin(u)we get I  =   
        (∫_0
              ^(π/2
                ))(((sin^n) (x))/((sin^n) (x)  +  
                    (cos^n) (x))) dx  =  
          -(∫_(π/2)^0)
                  (( (cos^n) (u))/((cos^n) (u)  +  
                        (sin^n) (u) )) du  = 
            (∫_0
                ^(π/2
                  ))(((cos^n) (u))/((sin^n) (u)  +  
                      (cos^n) (u) )) du
Therefore,
I  +  I  =   
        (∫_0
                ^(π/2
                  ))(((sin^n) (t))/((sin^n) (t)  +  
                      (cos^n) (t))) dt  + 
            (∫_0
                ^(π/2
                  ))(((cos^n) (t))/((sin^n) (t)  +  
                      (cos^n) (t))) dt  =  
          (∫_0
                ^(π/2
                  ))(((sin^n) (t)  +  
                      (cos^n) (t))/((sin^n) (t)  +  
                      (cos^n) (t))) dt  =  
            (∫_0^(π/2))dt =  
              (π/2)
and I  =  (π/4).