1
As a preliminary step we compute
lim
x→∞
x + sin((x)
sin(x)
= lim 1 +
= 1 + 0 = 1.
x→∞
x
x
Since arctan is a continuous function we get
π
x + sin(x)
= arctan(1) =
lim arctan
x→∞
x
4
Answer:
π
4
2
We compute the derivative wrt x of both sides and find that
y 0 · (1 + tan2 (y)) = 1 + y 0 .
This is equivalent to
y 0 · tan2 (y) = 1.
(1)
Since we know that tan(y) = x + y we insert this into our equation (1). We
find that
y 0 · (x + y)2 = 1,
or
y0 =
Answer: y 0 =
1
(x + y)2
1
(x+y)2 .
3
We use the substitution x = u2 . We find that
of integration are 0 and 4. Thus,
Z
0
4
√
x
dx =
x+1
Z
0
2
2u2
du = 2
u2 + 1
Answer: 4 − 2 arctan(2).
Z
0
2
dx
du
= 2u and that the new limits
Z 2
u2 + 1
1
du − 2
du =
2+1
u2 + 1
u
0
= 4 − 2[arctan(u)]20 = 4 − 2 arctan(2).
4
An integrating factor is esin(x) . We multiply the equation by this factor and
rewrite.
esin(x) y 0 + esin(x) cos(x)y = 0 ⇔ (esin(x) y)0 = 0 ⇔ esin(x) y = C ⇔ y = Ce− sin(x) .
Answer: y = Ce− sin(x) , where C is an arbitrary constant.
5
We apply the ratio test. Set an = n2 xn . Then
(n + 1)2
1
|an+1 |
= lim
· |x| = lim (1 + )2 · |x| = 12 · |x| = |x|.
2n
n→∞
n→∞
n→∞ |an |
n
lim
By the ratio test the series converges if |x| < 1 and diverges if |x| > 1.
Answer: The radius of convergence is 1.