Science:Math Exam Resources/Courses/MATH101/April 2015/Question 06 (a)

MATH101 April 2015

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[hide] Question 06 (a)
Find the Maclaurin series for $\int x^{4}\arctan(2x)dx$.

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[show] Hint
Find the Maclaurin series of $\arctan(x)$, and try to use that to write the series for the integrand.

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[show] Solution

Let $f(x)=\arctan(x)dx$, we start by finding the Maclaurin series for $f(x)$. One way is by finding the 1st, 2nd, 3rd, ... derivatives of $f(x)$ and at each step evaluate $f^{(n)}(0)$, and write the series. However, to avoid the cumbersome computation of the differentiation we use the fact that $f'(x)={\frac {1}{1+x^{2}}}$ i.e. $\arctan(x)$ is the anti-derivative of ${\frac {1}{1+x^{2}}}$, so $\arctan(x)=\int {\frac {1}{1+x^{2}}}dx$. On the other hand, we know that $\sum _{n=0}^{\infty }x^{n}={\frac {1}{1-x}}$. If we take $y=-x^{2}$, then we have ${\frac {1}{1+x^{2}}}={\frac {1}{1-(-x^{2})}}={\frac {1}{1-y}}=\sum _{n=0}^{\infty }y^{n}=\sum _{n=0}^{\infty }(-x^{2})^{n}=\sum _{n=0}^{\infty }(-1)^{n}x^{2n}$ Therefore, ${\begin{aligned}\arctan(x)=\int {\frac {1}{1+x^{2}}}dx&=\int \left(\sum _{n=0}^{\infty }(-1)^{n}x^{2n}\right)dx\\&=\sum _{n=0}^{\infty }\int (-1)^{n}x^{2n}\\&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {x^{2n+1}}{2n+1}}\\&=x-{\frac {x^{3}}{3}}+{\frac {x^{5}}{5}}-\cdots \end{aligned}}$ From this series we now build the series for the original function: $\arctan(2x)=\sum _{n=0}^{\infty }(-1)^{n}{\frac {(2x)^{2n+1}}{2n+1}}=2x-{\frac {(2x)^{3}}{3}}+{\frac {(2x)^{5}}{5}}-\cdots =2x-{\frac {8x^{3}}{3}}+{\frac {32x^{5}}{5}}-\cdots$ $x^{4}\arctan(2x)=\sum _{n=0}^{\infty }(-1)^{n}2^{2n+1}{\frac {x^{2n+5}}{2n+1}}=2x^{5}-{\frac {8x^{7}}{3}}+{\frac {32x^{9}}{5}}-\cdots$ ${\begin{aligned}\int x^{4}\arctan(2x)dx&=\int \left(\sum _{n=0}^{\infty }(-1)^{n}2^{2n+1}{\frac {x^{2n+5}}{2n+1}}\right)dx\color {blue}=\sum _{n=0}^{\infty }(-1)^{n}2^{2n+1}{\frac {x^{2n+6}}{(2n+1)(2n+6)}}\\&\color {blue}={\frac {1}{3}}x^{6}-{\frac {1}{3}}x^{8}+{\frac {16}{25}}x^{10}-\cdots \end{aligned}}$