Approximating Integrals within an error with Maclaurin Series "Use Maclaurin's Series to approximate the integral to 3dp accuracy"
$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}} $$
I was wondering if it is possible to solve this question by identifying out the number of terms needed to obtain a 3dp accuracy. I thought of using the Taylor Series Remainder Theorem, but it seems quite difficult with this binomial expression. 
 A: Yes it is possible:
The MacLaurin expansion of $$(1+x^2)^{-1/4}=1-\frac{x^2}{4}+\frac{5}{32}x^4+...$$
The numerical value of $$I=\int_{0}^{1/2} \frac{dx}{(1+x^2)^{1/4}}=0.490447$$
and $$\int_{0}^{1/2} (1-\frac{x^2}{4}+\frac{5}{32}x^4) dx= 0.409056.$$
A: Using the binomial expansion, we have
$$ \frac{1}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \binom{-\frac{1}{4}}{n} x^{2 n}$$ which, integrated gives
$$\int_0^{1/2} \frac{dx}{\sqrt[4]{x^2+1}}=\sum_{n=0}^\infty \frac{2^{-(2 n+1)}}{2 n+1} \binom{-\frac{1}{4}}{n}$$
So, summing $p$ terms, you look for $p+1$ such that
$$R_p=\Big| \frac{2^{-(2 p+3)}}{2 p+3} \binom{-\frac{1}{4}}{p+1}\Big| \leq \epsilon$$ 
Using the representation in terms of the gamma function we have
$$R_p=\frac{2^{-(2 p+3)}}{2 p+3} \frac{\Gamma \left(\frac{3}{4}\right)}{\Gamma (p+2) \left|\Gamma   \left(-p-\frac{1}{4}\right)\right|}$$ and using the reflection formula this becomes
$$R_p=\frac{2^{-(2 p+3)} \Gamma \left(\frac{3}{4}\right) \Gamma \left(p+\frac{5}{4}\right)
   \left|\sin \left(\left(p+\frac{5}{4}\right) \pi \right)\right|}{\pi  (2 p+3)
   \Gamma (p+2)}=\frac{2^{-(2 p+\frac{7}{2})} \Gamma \left(\frac{3}{4}\right) \Gamma
   \left(p+\frac{5}{4}\right)}{\pi  (2 p+3) \Gamma (p+2)}$$ Now, using Stirling approximations and continuing with Taylor series to $O\left(\frac{1}{p}\right)$,we have to solve for $p$
$$-2 p \log (2)-\frac{7 \log (p)}{4}+\log \left(\frac{\Gamma
   \left(\frac{3}{4}\right)}{16 \sqrt{2} \pi }\right)=\log(\epsilon)$$ and the solution is given in terms of Lambert function
$$p=\frac{7 }{8 \log (2)}W\left(k \right)\qquad \text{where} \qquad k=\frac{2^{3/7} \log (2)}{7} \left(\frac{\Gamma \left(\frac{3}{4}\right)}{16 \sqrt{2} \pi  \epsilon }\right)^{4/7}$$ Numerically, this gives
$$p\sim 1.26236 \, W\left(\frac{0.0130926 } { \epsilon^{4/7}}\right)$$ Computed for $\epsilon=10^{-10}$, this would give $p=8.70$ while the exact solution would be $p=8.76$.
