Recall that ∫ 1/(1+x^2) dx = arctan(x)
so integrate the series for 1/(1+x^2) term by term
Use the series for f of x equals 1 over the quantity 1 plus x squared to write the series for g(x) = tan–1(x).
A. C plus x minus x cubed over 3 plus x to the 5th power over 5 minus ...
B. C plus x minus x squared over 2 plus x cubed over 3 minus ...
C. C + 2 – 2x – 3x2 –
D. None of these
5 answers
so B?
stop guessing and look up the series involved to confirm your work.
sorry i'm not guessing i'm just confused and need help.
1/(1+x^2) = 1 - x^2 + x^4 - x^6 + ...
integrating that, you get
∫ 1/(1+x^2) dx = x - x^3/3 + x^5/5 - x^7/7 + ...
Now, if you check the series for arctan(x), you get
arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
ta-daa! choice A
Now, why does it say C at the front?
integrating that, you get
∫ 1/(1+x^2) dx = x - x^3/3 + x^5/5 - x^7/7 + ...
Now, if you check the series for arctan(x), you get
arctan(x) = x - x^3/3 + x^5/5 - x^7/7 + ...
ta-daa! choice A
Now, why does it say C at the front?
3 answers