[tex] \bold{PROBLEM:}[/tex]
[tex] \begin{aligned} \bold{ \int_{0}^{π/2} \cos {}^{2n} (x)dx.} \end{aligned}[/tex]
[tex] \large \bold{SOLUTION:}[/tex]
[tex] \bold{I \: try \: this. \: Notice \: that,}[/tex]
[tex] \begin{aligned} \bold{ \cos {}^{2n} x = \bigg ( \frac{e {}^{ix} + e {}^{ - ix} }{2} \bigg) {}^{2n} } = \bold{ \frac{1}{2 {}^{2n} } \sum_{k = 0}^{2n} \bigg( \frac{2n}{k} \bigg )e {}^{ikx} e {}^{ - i \bigg(2n - k \bigg)x} } \\ \\ \bold{ = \frac{1}{2 {}^{2n} }^{} } \sum_{k = 0}^{2n}\bold{ \bigg( \frac{2n}{k} \bigg )e {}^{ - i \bigg(2k- n \bigg)x} }\end{aligned}[/tex]
[tex] \bold{The\: terms \:with\:k\ne n } \\ \bold{integrate \: to \: zero \: over} \\ \bold{[0,\pi/2] \: and \: we \: are \: left \: with}[/tex]
[tex] \begin{aligned} \bold{ \int_{0}^{\pi/2}cos {}^{2n}x \: dx = \int_{0}^{ \pi/2 }} \bold{ \frac{1}{2n} \bigg( \frac{2n}{n} \bigg)dx = \frac{\pi}{2 {}^{2n + 1} } \bigg ( \frac{2n}{n} \bigg ) } \end{aligned}[/tex]
Another Solution because im not sure to my answer.
[tex] \bold{A \: given \: component \: e^{i(2k-2n)x}} \\ \bold{does \: not \: necessarily \: have}[/tex]
[tex] \begin{aligned} \bold{\int_{0}^{\frac{\pi}{2}}e^{i(2k-2n)x} dx = 0} \end{aligned}[/tex]
[tex] \bold{e.g. take k=n+1. However, \: we \: in \: fact \: have \: that}[/tex]
[tex] \begin{aligned} \bold{\int_{0}^{\frac{\pi}{2}}e^{-2kix} + e^{2kix}dx = 2\int_{0}^{\frac{\pi}{2}} \cos\left(2kx\right)dx=0} \end{aligned}[/tex]
and this is why your proof holds and symmetry of the binomial coefficient
OR MAYBE LIKE THIS
[tex] \begin{aligned} \bold{\int_0^{\pi/2} \exp(2 i x) d x = i \neq 0.} \end{aligned}[/tex]
I made everything and that's all I can do
