Using power series representation find an approximation to $\int_{0}^{0.1} \arctan(2x)dx$ Using power series representation find an approximation to 
$$\int_{0}^{0.1} \arctan(2x)dx$$
This is my solution:
$$\frac{d}{dx}\arctan(2x)=\ \frac{2}{1+4x^2} $$
$$\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$
$$\arctan(2x)=\ \int_{0}^{0.1}\frac{2}{1+4x^2} dx$$
$$= \int_{0}^{0.1}2\frac{1}{1-(-4x^2)} \, dx =\int_{0}^{0.1} \sum_{n=0}^\infty 2(-1)^n (4x^2)^n \, dx= \sum_{n=0}^{\infty}\frac{2(-1)^{n}4^{n}0.1^{2n+1}}{2n+1}$$
When I input the formula that I got in the calculator I get a value of 0.1973. But when I input the integral of arctangent I get 0.0099. I guess I missed something and can't find what. Where is my mistake?
 A: No, you are confusing things and you are not doing what you are required to do: you are not required to integrate the derivative of $\arctan$, but $\arctan$ itself.
Start by finding the Taylor series of $\arctan x$ around $0$:
$$(\arctan x)' = \frac 1 {1 + x^2} = \sum _{n = 0} ^\infty (-x^2)^n \implies \arctan x = \int \sum _{n = 0} ^\infty (-x^2)^n \ \Bbb d x = \sum _{n = 0} ^\infty (-1)^n \frac {x^{2n+1}} {2n+1} + C .$$
Since $\arctan 0 = 0$, we get that $C=0$.
Next, plugging $2x$ instead of $x$ in the above series leads to
$$\arctan 2x = \sum _{n = 0} ^\infty (-1)^n 2^{2n+1} \frac {x^{2n+1}} {2n+1} .$$
Using this series we may write
$$\int \limits _0 ^{0.1} \arctan 2x \ \Bbb d x = \int \limits _0 ^{0.1} \sum _{n = 0} ^\infty (-1)^n 2^{2n+1} \frac {x^{2n+1}} {2n+1} \ \Bbb d x = \sum _{n = 0} ^\infty \frac {(-1)^n 2^{2n+1}} {2n+1} \int \limits _0 ^{0.1} x^{2n+1} \ \Bbb d x = \\
\sum _{n = 0} ^\infty \frac {(-1)^n 2^{2n+1}} {2n+1} \frac {x^{2n+2}} {2n+2} \Bigg| _0 ^{0.1} = \sum _{n = 0} ^\infty \frac {(-1)^n 2^{2n+1}} {2n+1} \frac {(0.1)^{2n+2}} {2n+2} = \\
0.02 \sum _{n = 0} ^\infty \frac {(-0.04)^n} {(2n+1)(2n+2)} .$$
Since the terms of this series decrease very quickly, it is enough to compute just the first 3 of it in order to obtain a satisfactory approximation:
$$0.02 \sum _{n = 0} ^2 \frac {(-0.04)^n} {(2n+1)(2n+2)} = 0.02 \left( \frac 1 2 - \frac {0.04} {12} + \frac {0.04^2} {30} \right) = 0.0099344 .$$
A: The Taylor series (for $|x|\leq 1$) is $$\tan^{-1}(x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}x^{2n+1}$$ So  $$\tan^{-1}(2x)=\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}2^{2n+1}x^{2n+1}$$ Integrating $$\int \tan^{-1}(2x)\,dx=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)(2n+2)}2^{2n+1}x^{2n+2}$$ Using the bounds $$\int_0^{\frac 1{10}} \tan^{-1}(2x)\,dx=\frac{1}{100}-\frac{1}{15000}+\frac{1}{937500}-\frac{1}{43750000}+\cdots$$
A: You need to obtain a power series for $\arctan 2x$ and integrate that. What you have done is to obtain a power series for its derivative and integrate that, which is clearly wrong.
A: We have:
$$ I=\int_{0}^{\frac{1}{10}}\arctan(2x)\,dx = \frac{1}{2}\int_{0}^{\frac{1}{5}}\arctan(x)\,dx \stackrel{IBP}{=} \frac{\arctan\frac{1}{5}}{10}-\frac{\log\left(1+\frac{1}{25}\right)}{4}\tag{1}$$
where, for any $x\in(0,1)$,
$$ \arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{(2n+1)},\qquad \log(1+x^2)=\sum_{n\geq 1}\frac{(-1)^{n+1} x^{2n}}{n}.\tag{2} $$
Due to Leibniz' test,
$$ \left|\arctan\frac{1}{5}-\sum_{n=0}^{1}\frac{(-1)^n}{5^{2n+1}(2n+1)}\right|<\frac{1}{5^6},\qquad \left|\log\left(1+\frac{1}{25}\right)-\sum_{n=1}^{2}\frac{(-1)^{n+1}}{5^{2n}n}\right|<\frac{1}{3\cdot 5^6} $$
hence it follows that:
$$ I \approx \frac{1}{10}\left(\frac{1}{5}-\frac{1}{3\cdot 5^3}\right)-\frac{1}{4}\left(\frac{1}{5^2}-\frac{1}{2\cdot 5^4}\right) = \color{red}{\frac{149}{15000}}=0.00993333\ldots \tag{3}$$
and the approximation error is less than $2\cdot 10^{-5}$.
A: All the calculation is right but the answer is for $[arctan(2x)]_{0}^{0.1}$ not for $\int_{0}^{0.1} \arctan(2x)dx$ since $\int_{0}^{0.1}\frac{d}{dx}\arctan(2x)=[arctan(2x)]_{0}^{0.1}$ By fundamental theorem of calculus.
To solve the question, try to use integration by parts. And the answer is $x \arctan(2x) - \frac{1}{4} log(4x^2+1)$ which can be written as:
$$ \sum_{k =0}^\infty \frac{1}{1+2k}(-1)^k 2^{1+2k}   x ^{2+2k} - \frac{1}{4} \frac{1}{k}(-4)^k   (x ^{2})^k   $$
Evaluating it on $x = 0.1$ and $x =0$.
