C:\Users\Hank Kuiper\Desktop\integralanswers.xps

int
1
,x
2
cos x
sin x
(1)
cos x
NOTE: You must supply the extra arbitrary constant in each case : + C
NOTE: MAPLE seems to have an aversion to the tangent function - so the above answer is better as tan
(x)+C
int
int
int
x
,x
x C1
4
x
2
x C1
1
arctan x2
2
(2)
1
ln x2 C1
2
(3)
,x
3$x C7
20
,x ;
1
63
int
3$x C7
3 x C7
21
K1
,x
1
ln 3 x C7
3
int
int
(4)
(5)
cos x
,x
1 Csin2 x
cos x
1 Csin x
2
arctan sin x
(6)
1
K
1 Csin x
(7)
,x
int x$exp x2 , x
1 x2
e
2
(8)
sin ln x
cos ln x
(9)
int xK1$sec2 ln x , x
int tan 5$x $sec2 5$x , x
1
sec 5 x
10
2
(10)
int x2$exp K2$x , x
1
4
K
2 x2 C2 x C1 eK2 x
(11)
int x$sin x , x
sin x Kx cos x
(12)
int x$arcsin x , x
1
1 2
x arcsin x C x
4
2
2
Kx C1 K
1
arcsin x
4
(13)
int x$exp a$x , x
ax
int exp 2$ln csc x
a x K1 e
2
a
(14)
cos x
sin x
(15)
,x
K
int
2
,x
x C4
2
arctan
int
1
x
2
(16)
2 Cx
,x
x Kx K2
2
1
4
ln x C1 C ln x K2
3
3
K
int
(17)
5
,x
x K4$x
2
5
5
ln x K4 K ln x
4
4
3 Cx
,x
x C4 $ x2 C1
1
1
1
1
ln x2 C1 Carctan x K ln x2 C4 K arctan
x
6
6
2
2
int
(18)
2
(19)
2
int x $arctan x , x
int
3 Cx
x$ 7$x C1
2
1 3
1 2
1
x arctan x K
x C ln x2 C1
3
6
6
(20)
20
7 7 x C1
(21)
,x
K3 ln 7 x C1 C3 ln x C
x4
int
,x
x K1 $ x K2 $ x K3
1 2
1
81
x C6 x C ln x K1 C
ln x K3 K16 ln x K2
2
2
2
2$x C9
int 2
,x
x C2$x C10
7
1
1
ln x2 C2 x C10 C arctan
xC
3
3
3
int
csc2 3$x
sqrt 4 Kcot2 3$x
,x
(22)
(23)
1
12
2
1
4 sin 3 x
2
5 cos 3 x K4
2
cos 3 x K1
arcsin
1
10
K1 Ccos 3 x
5 cos 3 x K4
K
2
cos 3 x C2 cos 3 x C1
(24)
5 5 cos 3 x C4
1 Ccos 3 x
2
5 cos 3 x K9 cos 3 x C4
Karctan
2
5 cos 3 x K4
sin 3 x
K
2
cos 3 x C2 cos 3 x C1
This shows you you can't always depend on math software to give you a decent answer. If we let u=cot
(3x)/2, then du=-(3/2)csc^2(3x) and the integral becomes
1
1
K
$ int
,u
2
3
sqrt 1 Ku
1
K arcsin u
(25)
3
so the answer is -(1/3) arcsin(cot(3x)/2)
cot 3$x
simplify arcsin
2
1
arcsin
cot 3 x
(26)
2
2
3
int exp x $sin exp x $cos exp x , x
1
x
sin e
4
int
1 C2$x C3$x2 C4$x3
2
2
4
(27)
,x
5$x C6
1 K130 x C168
23
C
2
600
1800
5 x C6
The last problem is the same as #19
30 arctan
1
x
6
30
C
2
2
ln 5 x C6
25
(28)