Inspired by Andrew Ellinor

Geometry Level 4

$\Large \mathfrak P=\displaystyle\int_0^{\pi/2}\mathfrak{T}(\phi)\mathrm{d}\phi$

Given $\large \mathfrak{T}(\phi)=\displaystyle\sum_{n=0}^{\infty}\dfrac{\cos^3\phi}{\csc^{4n} \phi},~(|\csc \phi|\neq 1)$ , if the value of $\mathfrak P$ can be written in the form $\large\dfrac{\pi}{\psi^{\psi}}$ $~~(\psi\in\mathbb{Z}^+)$ , then: $\Large\color{#D61F06}{\psi^{\psi^{\psi+1}}}=\ ?$

Inspiration

None of the other options 16 256 This question is flawed

2^5\times 2^2

1 solution

Rishabh Jain
May 5, 2016

$\displaystyle\sum_{n=0}^{\infty}\sin^{4n}\phi$ represent a sum of Infinite GP $~~(\because \sin^4 \phi<1)$ which is given by: $\dfrac{1}{1-\sin^4\phi}=\dfrac{1}{\underbrace{(1-\sin^2\phi)}_{\cos^2\phi}(1+\sin^2\phi)}$ Hence, $\Large\mathfrak{T}(\phi)=\dfrac{\cos \phi}{1+\sin^2\phi}$ $\begin{aligned}\therefore \mathfrak{P}=\displaystyle\int_0^{\pi/2}\dfrac{\overbrace{\cos \phi~\mathrm{d}\phi}^{\mathrm{d}(\sin \phi)}}{1+\sin^2\phi}&=\left|\tan^{-1}(\sin \phi)\right|_0^{\pi/2}\\&=\large \dfrac{\pi}{4}=\dfrac{\pi}{2^2}\end{aligned}$

$\Large \therefore 2^{2^3}=2^8=\huge\boxed{256}$

Inspired by Andrew Ellinor

1 solution

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