Download Solutions Calculus II - Spring 2013 - Written homework 6 1. Integrate

Solutions
Calculus II - Spring 2013 - Written homework 6
Z
1. Integrate (give your answer in terms of t):
√
1
dt
9 − t2
We should rewrite this one prior to substituting:
Z
Z
1
1
√
q
dt =
2
9−t
9 1−
Z
t2
dt =
9
1
q
3 1−
t 2
3
dt
= sin θ, so dt = 3 cos θ dθ, and, for back-substitution purposes, we have θ = arcsin 3t :
Z
Z
Z
1
3 cos θ dθ
cos θ
p
√
dθ = 1 dθ
=
3
cos2 θ
1 − sin2 θ
t
= θ + C = arcsin
+C
3
Z
3x
√
2. Integrate (give your answer in terms of x):
dx
x2 − 4
Now, let
t
3
Notice that this one could easily be done with a regular substitution! But, if you didn’t catch that, then
you could have proceeded with a trigonometric substitution, as follows. First, it must be rewritten a
bit:
Z
Z
Z
Z
3x
x
x
x
3
√
q
q dx
=
3
dx = 3 q
dx
=
dx
2
2
2
x2
x2 − 4
x
x
4 4 −1
2
−
1
−
1
2
2
Now it is clear that the substitution should be x2 = sec θ, whence x = 2 sec θ, dx = 2 sec θ tan θ dθ, and
for back-substitution purposes, we note that θ = arcsec x2 . Rewrite the integral to get:
3
2
Z
√
2 sec θ
2 sec θ tan θ dθ = 6
sec2 θ − 1
Z
sec2 θ tan θ
√
dθ = 6
tan2 θ
= 6 tan θ + C = 6 tan arcsec
Z
sec2 θ dθ
x
+C
2
Again, we use a triangle to recover tan arcsec x2 :
√
x2 − 4. Thus, the final answer becomes 3 x2 − 4 + C.
Z
x2
3. Integrate (give your answer in terms of x):
dx
3/2
(1 + x2 )
From this we calculate tan arcsec x2 =
1
2
√
Using trigonometric substitution, let x = tan θ. Recall for our later back substitution that this means
θ = arctan x. Then dx = sec2 θ dθ, and the integral becomes:
Z
Z
Z
Z
tan2 θ sec2 θ
tan2 θ sec2 θ
tan2 θ sec2 θ
tan2 θ
dθ
=
dθ
=
dθ
=
dθ
3/2
3/2
sec3 θ
sec θ
(sec2 θ)
1 + tan2 θ
1
Z
=
Z
=
sin2 θ
dθ =
cos θ
Z
sin2 θ
cos2 θ
1
cos θ
Z
dθ =
1 − cos2 θ
dθ =
cos θ
Z
cos θ
dθ
sin2 θ
1
− cos θ dθ =
cos θ
Z
sec θ − cos θ dθ
= ln | sec θ + tan θ| − sin θ + C = ln |sec arctan x + x| − sin arctan x + C
To recover an elementary value for x, sketch a right triangle with an angle θ whose tangent is x, find
the missing third side of the triangle, and use this to find the other values:
√
x
From this we get sin arctan x = √1+x
, and sec arctan x = x2 + 1, so the solution to the integral is
2
p
x
+ C.
ln x2 + 1 + x − √
x2 + 1
Z
7
dx
4. Integrate, using partial fraction decomposition:
(x − 1)(2x + 4)
We set up the sum as follows:
A
B
7
=
+
, whence
(x − 1)(2x + 4)
x − 1 2x + 4
7
2Ax + 4A + Bx − B
=
(x − 1)(2x + 4)
(x − 1)(2x + 4)
This gives us two equations to solve, 4A − B = 7, and 2A + B = 0. Solve these using elementary
algebra to get B = − 37 and A = 76 ; this yields the integral
Z
7
1
1
7
7 x − 1 −
= (ln |x − 1| − ln |x + 2|) + C = ln +C
6 x−1 x+2
6
6
x + 2
2