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