pi-2-0-dx-1-tan-2014-x-pie-q-p-Find-2p-q-

Question Number 133036 by liberty last updated on 18/Feb/21

{\int_{0}}^{\frac{π}{2}} \frac{d x}{1 + \tan^{2014} (x)} = \frac{π e^{q}}{p}

Find 2 p - q .

Answered by liberty last updated on 18/Feb/21

I = {\int_{0}}^{\frac{π}{2}} \frac{dx}{1 + \tan^{2014} (x)} = {\int_{\frac{π}{2}}}^{0} \frac{\tan^{2014} (x)}{1 + \tan^{2014} (x)} (- dx)

I = {\int_{0}}^{\frac{π}{2}} \frac{\tan^{2014} (x)}{1 + \tan^{2014} (x)} dx

we get 2 I = {\int_{0}}^{\frac{π}{2}} \frac{1 + \tan^{2014} (x)}{1 + \tan^{2014} (x)} dx

I = \frac{1}{2} . \frac{π}{2} = \frac{π}{4} = \frac{π e^{q}}{p}

\to {\begin{cases} q = 0 \\ p = 4 \end{cases} \Rightarrow 2 p - q = 8

Answered by mathmax by abdo last updated on 20/Feb/21

I = \int_{0}^{\frac{π}{2}} \frac{dx}{1 + \tan^{2014} x} \Rightarrow I =_{tanx = z} \int_{0}^{\infty} \frac{dz}{(1 + z^{2}) (1 + z^{2014})}

I =_{z = \frac{1}{t}} - \int_{0}^{\infty} \frac{- dt}{t^{2} (1 + \frac{1}{t^{2}}) (1 + \frac{1}{t^{2014}})} = \int_{0}^{\infty} \frac{t^{2014}}{(1 + t^{2}) (1 + t^{2014})} \Rightarrow

2 I = \int_{0}^{\infty} (\frac{1}{(1 + z^{2}) (1 + z^{2014})} + \frac{z^{2014}}{(1 + z^{2}) (1 + z^{2014})}) dz = \int_{0}^{\infty} \frac{dz}{1 + z^{2}}

= \frac{π}{2} \Rightarrow I = \frac{π}{4} = \frac{π}{p} e^{q} \Rightarrow p = 4 and q = 0 \Rightarrow 2 p - q = 8