Calculus exercise 8.1
6.
12.
R dx
5x
=
−1 −x
5
ln5
R e√x
√ dx
x
+c
√
= 2e
x
+c
13. u = 1/x, du = −x−2 dx
√
R 1/x
R 1/2
=⇒ 21 ex2 dx = 1 − eu du = e − e
18. u = R3 − 2cosφ
sinφ
=⇒ 3−2cosφ
dφ = 21 ln(3 − 2cosφ) + c
22.
R ln x
dx
25.
R
dx
x2 +6x+10
43.
R
1
dx
x(2x+3)
49.
R π√
1 + cosxdx
x
= 12 (ln x)2 + c
=
R
dx
(x+3)2 +1
= arctan x + 3 + c
x
= 13 ln | 2x+3
|+c
0
=
R πq
2cos2 ( x2 )dx =
0
√ Rπ
√
2 0 cos x2 dx = 2 2
50.
(a) Obviously
(b) Obviously
(c) Using the formula tan2 x + 1 = sec2 x determine the C1 and C2
51.
R π1
(a)
Rπ
sin2 nxdx
(b)
Rπ
sin nx cos nxdx
(c)
R
0
=
02
0
π
n
0
sin nx cos nx =
−
=
1
2
1
2
R
cos2nx
dx
2
π
2
=
Rπ
sin 2nxdx
0
π
n
0
=0
π/n
sin 2nxdx = −[ cos4n2nx ]0
=0
53.
(a)
3
tan
xdx = (tan2Rx)tanxdx = (sec2 x − 1)tanxdx
R
= sec2 xtanxdx − tanxdx
= 12 tan2 x − ln|secx| + c
(b)
R
R
R
R
5
2
tan
xdx = (tan3 x)tan
xdx = (sec2 x − 1)tan3 xdx
R
R
2
3
3
= sec xtan xdx − tan xdx
= 14 tan4 x − 12 tan2 x + ln|secx| + c
R
R
1
7
2
tan
xdx = (tan5 x)tan
xdx = (sec2 x − 1)tan5 xdx
R
R
2
5
5
= sec xtan xdx − tan xdx
= 61 tan6 x − 14 tan4 x + 12 tan2 x − ln|secx| + c
(c)
R
(d)
R
R
R
2k+1
tan
xdx = (tan2k−1
x)tan2 xdx = (sec2 x − 1)tan2k−1 xdx
R
R
= sec2 xtan2k−1
xdx − tan2k−1 xdx
R
1
2k
= 2k tan x − tan2k−1 xdx
R
R
2