Answers to
MATHEMATICA Assignment #1
Calculus 4, Section 005 


#1
Integrate[ Cos[x] / (1+Sin[x]^2), x ]
ArcTan[Sin[x]]

#2
Integrate[  Cos[x] / Sqrt[1+Sin[x]] , x ]
2 Sqrt[1 + Sin[x]]

#3
Integrate[ Tan[x]/Cos[x]^2 , x ]
      2
Sec[x]
-------
   2


#4
Integrate[ Tan[x]^3 , x ]
                    2
              Sec[x]
Log[Cos[x]] + -------
                 2

#5
Integrate[ Sin[x]^4 , x ]
12 x - 8 Sin[2 x] + Sin[4 x]
----------------------------
             32

#6
Integrate[ 1/Sqrt[4-x^2] , x ]
       x
ArcSin[-]
       2

#7
Integrate[ 1/(x + x^2) , x ]
Log[x] - Log[1 + x]

#8
Integrate[ x Cos[x]^2 , x ]
   2
2 x  + Cos[2 x] + 2 x Sin[2 x]
------------------------------
              8

#9
Integrate[ x ArcSin[x] , x ]
            2                 2
x Sqrt[1 - x ]   ArcSin[x]   x  ArcSin[x]
-------------- - --------- + ------------
      4              4            2

#10
Integrate[ Cos[Sqrt[x]] , x ]
2 (Cos[Sqrt[x]] + Sqrt[x] Sin[Sqrt[x]])

#11
Integrate[ ArcSin[x]/Sqrt[1-x^2] , x ]
         2
ArcSin[x]
----------
    2

#12
Integrate[  x Sec[x]^2, x ]
Log[Cos[x]] + x Tan[x]

When each of the above answers is differentiated with respect to x the original function will be obtained. This is easy to check for problems 1, 2, 10, 11, and 12 where the same form of the original expression is obtained when MATHEMATICA is asked to differentiate the answer. On the other hand, if we look at problem #5 as an example we find: 
#5
Integrate[ Sin[x]^4 , x ]
12 x - 8 Sin[2 x] + Sin[4 x]
----------------------------
             32

D[ %, x ] 
12 - 16 Cos[2 x] + 4 Cos[4 x]
----------------------------
             32
which initially looks different. However, if we continue we can check the answer by: 
Simplify[ % ]
      4
Sin[x]
The same approach works to check problems 4, 5, 6, 7, 8, and 9. For problem 3, the simplified answer is sec2(x)tan(x) but it is easy to check
sec2(x)tan(x) = sin(x)/cos3(x) = tan(x)/cos2(x)

This document was created on Feb. 4, 1997.

URL: http://math.ou.edu/~amiller/math/math1ans.htm