# Approximating an integral using Taylor Series

I have to approximate this integral with an error lesser than 0.1 using Taylor Series. This is the integral: $$\int_0^1 \arctan\left(\frac{1}{x^{10}}\right) \,\Bbb dx.$$ If I understood correctly, I have to determinate the Taylor Series expansion in terms of a series with alternating signs. I have to check if the series has a uniform convergence in [0,1] and then take the series sign outside the integral sign and integrate the $$x$$-dependent part. Once done, I have to estimate the error using Leibniz: $$|S - S_{n+1}|<|a_n|$$. To find it I just have to calculate $$n$$ terms and when I’ll find $$a_n<0.1$$ I can computer the series from 0 to $$n-1$$. Am I correct?

I think I figured out how to do this job but I really do not know how to determinate the Taylor series expansion… I tried to use the McLaurin’s $$\arctan(x)$$ where $$x_0=0$$ but the integrand function is not defined for $$x=0$$… how can I handle this kind of problem? Please, can someone explain me if everything I said is correct and explain me how to determinate the series expansion in this case?

• You might want to use $\arctan (1/u) = \pi/2 - \arctan u.$
– zhw.
May 25, 2016 at 17:04
• Can you please explain me where did you take that relationship? I have more exercises like this (exp(1/x) for example) and I'd like to know how to handle these problems. Thank you in advance May 27, 2016 at 20:18
• You can't expect this nice relation to hold too often; $e^{1/x}$ is quite a different situation.
– zhw.
May 27, 2016 at 20:55

$$I=\int_{0}^{1}\arctan\left(\frac{1}{x^{10}}\right)\,dx = \frac{\pi}{2}-\int_{0}^{1}\arctan(x^{10})\,dx$$ hence:
$$I = \frac{\pi}{2}-\sum_{n\geq 0}\frac{(-1)^n}{2n+1}\int_{0}^{1} x^{10+20n}\,dx = \color{red}{\frac{\pi}{2}-\frac{1}{11}+\frac{1}{93}}-\sum_{n\geq 2}\frac{(-1)^n}{(11+20n)(1+2n)}.$$
• That's fine, but be careful to round correctly. $\pi/2 - 1/11 = 1.48$, so to one decimal place you should answer 1.5. Jul 8, 2016 at 0:59