Freecalc
Homework on Lecture 6
Will be quizzed: time to be announced in class
1. Integrate.
Z
1
(a)
dx.
3 + cos x
x !
1
+C
√ tan
2
2
1
dx.
3 + sin x
Z
1
dx. (Hint: this integral can be done simply with the substitution x = arctan t.)
2 + tan x
1
answer: √ arctan
2
Z
(b)


3 tan x + 1
1
2

answer: √ arctan 
√
2
2 2
5
answer:
5
ln (sin x + 2 cos x) +
1
(c)
x+C
2
Solution. 1.a This integral is of none of the forms that can be integrated quickly. Therefore we have to use the
standard rationalizing substitution x = 2 arctan t, t = tan x2 . We recall that from the double angle formulas it
follows that
cos2 (arctan t) − sin2 (2 arctan t)
1 − t2
cos(2 arctan t) =
=
.
1 + t2
cos2 (arctan t) + sin2 (arctan t)
Therefore we can solve the integral as follows.
Z
Z
1
1
dx =
d (2 arctan t)
3 + cos x
3
+
cos(2
arctan t)
Z
1
2
=
dt
1−t2 (1 + t2 )
3 + 1+t2
Z
2
=
dt
Z 4 + 2t2
1
=
dt
2
+
t2
√
√ !
2
2
=
arctan
t +C
2
2
!
√
√
x
2
2
=
arctan
tan
+C .
2
2
2
Substitute x = 2 arctan t
Solution. 1.c This integral is of none of the forms that can be integrated quickly. Therefore we can solve it using the
standard rationalizing substitution x = 2 arctan t, t = tan x2 . This results in somewhat long computations and we
invite the reader to try it.
1
However, as proposed in the hint, the substitution x = arctan t works much
Z
Z
1
1
dx =
d (arctan t)
2 + tan x
2
+
tan(arctan
t)
Z
1
1
=
dt
(2
+
t)
(1
+
t2 )
Z 1
− 5t + 25
5
=
+
dt
(t + 2) (t2 + 1)
1
1
2
=
ln(t + 2) −
ln(t2 + 1) + arctan t + C
5
10
5
1
1
2
=
ln(tan x + 2) −
ln(tan2 x + 1) + x + C
5
10
5
1
1
2
=
ln(tan x + 2) + ln(cos x) + x + C
5
5
5
1
2
=
ln ((tan x + 2) cos x) + x + C
5
5
1
2
=
ln (sin x + 2 cos x) + x + C.
5
5
faster:
Substitute x = arctan t
Decompose into partial fractions
Substitute back t = tan x
2. Integrate. The answer key has not been proofread, use with caution.
Z
(a)
sin(3x) cos(2x)dx.
answer: − 1 cos(5x) − 1 cos x + C
10
2
Z
(b)
sin x cos(5x)dx.
answer: − 1 cos(6x) + 1 cos(4x) + C
12
8
Z
(c)
cos(3x) sin(2x)dx.
answer: − 1 cos(5x) + 1 cos x + C
10
2
Z
(d)
sin(5x) sin(3x)dx.
answer: 1 sin(2x) − 1 sin(8x) + C
4
16
Z
(e)
cos(x) cos(3x)dx.
answer: 1 sin(4x) + 1 sin(2x) + C
8
4
3. Integrate.
Z
(a)
sin2 x cos xdx.
answer: 1 sin3 x + C
3
Z
(b)
sin2 xdx.
answer: x − 1 sin(2x) + C
2
4
Z
(c)
cos3 xdx.
answer: sin x − 1 sin3 x + C
3
4. Integrate
Z
(a)
sec3 xdx.
answer: 1 (sec x tan x + ln | sec x + tan x|) + C
2
Z
(b)
tan3 xdx.
answer: 1 tan2 x − ln | sec x| + C
2
sec2 x tan2 xdx.
3
answer: tan x + C
3
Z
(c)
2