find polynominal of degree $3$ that approximates the function

Question

Question is:

Find the polynominal of degree $3$ that approximates the function $$ h(x)= \int \frac{\sqrt{1-4x^2}}{1+3x}\,\mathrm dx$$ where $a= \dfrac14$ and $b=0$.

I have already worked out the binominal series of $$(1-4x^2)^{1/2}$$ and $$(1+3x)^{-1}$$ as that was one of my previous questions, but I'm stuck now.

Answer 1

Since you already worked the binomial expansion $$\sqrt{1-4 x^2}=1-2 x^2-2 x^4-4 x^6+O\left(x^8\right)$$ just perform the long division by $(1+3x)$. This would give $$\frac{\sqrt{1-4 x^2}}{1+3 x}=1-3 x+7 x^2-21 x^3+61 x^4-183 x^5+545 x^6-1635 x^7+O\left(x^8\right)$$ Integrate termwise and stop at the degree you are asked for. Do not forget the integration constant.

Answer 2

Expand,

$$\sqrt{1-4x^2} = 1-2x^2+O(x^4), \>\>\>\>\> \frac1{1+3x} = 1-3x+9x^2+O(x^3)$$

to approximate the integrand,

$$\frac{\sqrt{1-4x^2}}{1+3x} = (1-2x^2+O(x^4))(1-3x+9x^2+O(x^3))=1-3x+7x^2+O(x^3)$$

Then,

$$h(x)= \int \frac{\sqrt{1-4x^2}}{1+3x} \approx \int (1-3x+7x^2)dx = x-\frac32x^2 +\frac73 x^3$$