Integrate the following (below) using infinite #\bb\text(SERIES)#?
#\int(4x^3)/(1+x^8)dx#
(If there are trig substitution steps, please explain if possible -- I'm pretty bad at those)
(If there are trig substitution steps, please explain if possible -- I'm pretty bad at those)
2 Answers
# int \ (4x^3)/(1+x^8) \ dx = 4x - (1)/(3)x^12 + (1)/(5)x^20 - (1)/(7)x^28 + ... + C#
Explanation:
We seek:
# I = int \ (4x^3)/(1+x^8) \ dx #
in the form of a Power Series. We can write the integral in the form:
# I = int \ 4x^3(1+x^8)^(-1) \ dx #
And we can utilise the Binomial Theorem:
# (1+x)^n = 1 + nx + (n(n-1))/(2!)x^2 + (n(n-1)(n-2))/(3!)x^3 + ...#
So then we can write:
# I = int \ (4x^3){1+(-1)(x^8) + ((-1)(-2))/(2!)(x^8)^2 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + ((-1)(-2)(-3))/(3!)(x^8)^3 + #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ + ((-1)(-2)(-3)(-4))/(4!)(x^8)^4 + ... }#
# \ \ = int \ (4x^3){1-x^8 + x^16 - x^24 + ... } #
# \ \ = int \ 4 - 4x^11 + 4x^19 - 4x^27 + ... #
# \ \ = 4x - (4)/(12)x^12 + (4)/(20)x^20 - (4)/(28)x^28 + ... + C#
# \ \ = 4x - (1)/(3)x^12 + (1)/(5)x^20 - (1)/(7)x^28 + ... + C #
Note:
In this particular case we can evaluate the integration by direct substitution, Let
# u = x^4 => (du)/dx = 4x^3 #
And if we substitute into the integral, we gtre:
# I = int \ 1/(1+u^2) \ du #
Which is a standard integral, so we get:
# I = arctanu+C #
And restoration of the substitution gives:
# I = arctan(x^4) + C#
And it can readily be verified that this solution has the above Power Series expansion.
See below.
Explanation:
Here's a solution without the Binomial Theorem, and rather just using the Power Series
As
Multiply in
Integrate term-by-term (don't forget constant of integration):
Now, recalling that