8.8. Calculations with Series http://www.ck12.org

ex

x =

e·eu

1 +u=e·

1

1 +u

∞

n∑= 0

(u)n

n!

=e( 1 −u+u^2 −u^3 +...)

(

1 +u+u

2

2!+

u^3

3!+

u^4

2!+...

)

=e

(

1 +^12 u^2 −^13 u^3 +^38 u^4 −^1130 u^5 +...

)

for|u|< 1

so
∫ex
xdx=e

(u+ 1

6 u^3 − 121 u^4 + 403 u^5 − 18011 u^6 +...

)whereu=x− 1

Example 2Find the power series representation of∫sinxx^2 dx
Solution. Direct substitution ofx^2 in the Maclaurin Series of sinxgives

sinx^2 =∑∞n= (^0) ( 2 (−n+^1 ) 1 n)!(x^2 )^2 n+^1 andsinxx^2 =∑n∞= (^0) ( 2 (−n+^1 ) 1 n)!x^4 n+^1
So
∫ sinx 2
x dx=
∞
n∑= 0
(− 1 )n
( 2 n+ 1 )!
∫
x^4 n+^1 dx

∞
n∑= 0
(− 1 )n
( 2 n+ 1 )!
x^4 n+^2
4 n+ 2 =
∞
n∑= 0
(− 1 )n
( 4 n+ 2 )
x^4 n+^2
( 2 n+ 1 )!
Exercises

1. Find the power series representation (Maclaurin Series) of∫ex^2 dxand approximate
   ∫^1
   0

ex^2 dxto 6 decimal

places.

2. Find the power series representation (Maclaurin Series) of∫sinx^2 dx.

Frequently Used Maclaurin Series

Some frequently used Maclaurin Series are listed below

1 −^1 x=∑∞n=^0 xn in(−^1 ,^1 )

( 1 −^1 x)^2 =∑∞n=^1 nxn−^1 in(−^1 ,^1 )

ex=∑∞n= 0 xnn! in(− 1 , 1 )

ln( 1 −x) =−∑∞n= 1 xnn, in(− 1 , 1 )

ln( 1 +x) =∑∞n= 1 (− 1 )n+^1 xnn, in(− 1 , 1 )

sinx=∑n∞= 0 (−( 21 n)n+x 12 n)+!^1 in(− 1 , 1 )

cosx=∑n∞= 0 (−(^12 )nn)x!^2 n in(− 1 , 1 )

tan−^1 x=∑∞n= 0 (−(^12 )nn+x^21 n)+^1 in(− 1 , 1 )