# Integration Worksheet with Solution - PDF by vsp19127

VIEWS: 0 PAGES: 3

• pg 1
```									                      Integration Worksheet
Math 132
Solutions

(a) Use partial fractions:
x+3           1      10         1   1
2 + 5x − 2
dx =             dx +          dx
3x                 7    3x − 1       7 x+2
10              1
=    ln(3x − 1) + ln(x + 2) + C.
21              7
(b) Use integration by parts with u = x, dv = e3x dx (which gives du = dx and
1
v = 3 e3x ):
1                     1               1
x      x                 1
3x
dx = e3x          −               e3x dx
0       e       3         0       3   0
1                   1
x          1
= e3x − e3x
3     0    9                   0
3
1 + 2e
=          .
9
√
(c) Use u substitution. Let u = x + 1 to get u2 = x + 1, 2udu = dx, x = u2 − 1.
Also, x = 0 ⇒ u = 1 and x = 8 ⇒ u = 3.
8 √            u=3
x x + 1 dx =      (u2 − 1)(u)(2udu)
0                       u=1
3
2u3 2u5
=−       +
3     5                   1
1192
=      .
15
1−cos(2t)
(d) Use the “half-angle” formula from trigonometry to write sin2 (t) =               2
.
1 − cos(2t)
sin2 (t) dt =              dt
2
1     1
= t − sin(2t) + C.
2     4
(e) Here, use the fact that cos2 (t) = 1 − sin2 (t):

cos3 (t) dt =       cos(t)(1 − sin2 (t)) dt

sin3 (t)
= sin(t) −            + C.
3

(f) Here, I recommend using the substitution x = sinh(t). Then, x2 + 1 = sin2 (t) +
1 = cosh2 (t) and dx = cosh(t)dt. (I prefer the hyperbolic trig substitution here
to the trigonometric substitution x = tan(t) because, in the end, I think it’s
harder to deal with sec3 (t) than cosh2 (t).)
√
x2 + 1 dx =              cosh2 (t) cosh(t) dt

=             cosh2 (t) dt
e2x + 2 + e−2x
=                     dt
4
e2t + 4t − e−2t
=                 +C
8

To ﬁnish, and put things in terms of x again, use the fact that x = sinh(t) ⇒
√
t = ln(x + 1 + x2 ) and a little algebra:

√                          √                              √
(x +       1 + x2 )2 + 4 ln(x +       1 + x2 ) − (x +                1 + x2 )−2
=                                                                                          +C
√             √     8
x 1 + x2 ln(x + 1 + x2 )
=         +                + C.
2           2

(g) Use integration by parts with u = arcsin(x) and dv = dx (then we have du =
√ dx  and v = x).
1−x2

1                                     1           1
x dx
arcsin(x) dx = x arcsin(x)            −           √
0                                         0       0           1 − x2
1       √              1
= x arcsin(x)         +       1 − x2
0                      0
π
= − 1.
2

2
(h) Here, we use parts with u = (ln(x))2 and dv = dx (then we have du = 2 ln(x) dx
x
and v = x).
e                                      e               e
2                     2
(ln(x)) dx = x(ln(x))                  −               2 ln(x) dx
1                                          1           1
e                                e
2
= x(ln(x))            + 2x − 2x ln(x)
1                                1
= e − 2.

1
(i) Here use the fact that ln x 2                = 1 ln(x):
2

√           1
ln           x dx =    ln(x) dx
2
x x ln(x)
=− +          + C.
2       2

(j) Here you have some choices. You could use the substitution x = 2 cosh(t).
This time, we’ll use the trigonometric substitution x = 2 sec(t). This leads
√
to x2 − 4 = 2 tan(t) and dx = 2 sec(t) tan(t)dt. Let’s ignore the limits of
√
20   dx
integration in √5 x2 √x2 −4 at ﬁrst.

dx                      2 sec(t) tan(t) dt
√       =
x2 x2 − 4                   8 sec2 (t) tan(t)
1
=    cos(t) dt
4
1
= sin(t) + C.
4
Probably the tidiest way to ﬁnish, is to use the fact that when x = 2 sec(t) =
√                                      √
1                                         2
5 ⇒ sin(t) = √5 and x = 2 sec(t) = 20 ⇒ sin(t) = √5 . Then, the deﬁnite
integral
√                                          √
x= 20                                      x= 20
dx      1                                        1           2   1              1
√       = sin(t)                              =               √ −√             = √ .
√
x= 5     x2 x2 − 4  4                         √
x= 5             4            5   5            4 5

3

```
To top