Science:Math Exam Resources/Courses/MATH307/December 2012/Question 04 (e)/Solution 1

Using the hint given in the question, we have

${\begin{aligned}\int _{0}^{1}\cos(2\pi t)(t^{2}-t)dt&=\int _{0}^{1}{\frac {1}{2}}(e^{2\pi it}+e^{-2\pi it})(t^{2}-t)dt\\&=\int _{0}^{1}{\frac {1}{2}}e^{2\pi it}(t^{2}-t)dt+\int _{0}^{1}{\frac {1}{2}}e^{-2\pi it}(t^{2}-t)dt\\&={\frac {1}{2}}\int _{0}^{1}(t^{2}-t)e^{2\pi it}dt+{\frac {1}{2}}\int _{0}^{1}(t^{2}-t)e^{-2\pi it}dt\\&={\frac {1}{2}}\langle t^{2}-t,e^{-2\pi it}\rangle +{\frac {1}{2}}\langle t^{2}-t,e^{2\pi it}\rangle \\&={\frac {1}{2}}c_{-1}+{\frac {1}{2}}c_{1}\end{aligned}}$

From part (d) we know that $c_{n}={\frac {1}{2\pi ^{2}n^{2}}}$, which we plug into our equation above to calculate

${\begin{aligned}\int _{0}^{1}\cos(2\pi t)(t^{2}-t)dt&={\frac {1}{2}}\left({\frac {1}{2\pi ^{2}(-1)^{2}}}\right)+{\frac {1}{2}}\left({\frac {1}{2\pi ^{2}(1)^{2}}}\right)\\&={\frac {1}{4\pi ^{2}}}+{\frac {1}{4\pi ^{2}}}={\frac {1}{2\pi ^{2}}}\end{aligned}}$