Computing $\int {\dfrac{\csc^{2014}x-2014}{\cos^{2014}x} dx}$

Question

I don't know how to compute:

$$\int {\dfrac{\csc^{2014}x-2014}{\cos^{2014}x} dx}$$

I have tried substituting $t=\tan ^{2} x$ but got nothing out of it. I know there's some trick involved, but can't figure it out.

Also, how does one frame such questions involving numbers like the current year, next year or previous year?

Is there a general theme to attack such problems?

Answer 1

This question seems to examine you how to use well the $\sum$ sign.

\begin{align} \int\dfrac{\csc^{2014}x-2014}{\cos^{2014}x}dx&=\int\dfrac{1}{\sin^{2014}x\cos^{2014}x}dx-\int\dfrac{2014}{\cos^{2014}x}dx\\ &=\int\dfrac{1}{\dfrac{\sin^{2014}2x}{2^{2014}}}dx-\int\dfrac{2014}{\cos^{2014}x}dx\\ &=\int2^{2014}\csc^{2014}2x~dx-\int2014\sec^{2014}x~dx\\ &=-\int2^{2013}\csc^{2012}2x~d(\cot2x)-\int2014\sec^{2012}x~d(\tan x)\\ &=-\int2^{2013}(1+\cot^22x)^{1006}~d(\cot2x)-\int2014(1+\tan^2x)^{1006}~d(\tan x)\\ &=-\int2^{2013}\sum\limits_{n=0}^{1006}C_n^{1006}\cot^{2n}2x~d(\cot2x)-\int2014\sum\limits_{n=0}^{1006}C_n^{1006}\tan^{2n}x~d(\tan x)\\ &=-\sum\limits_{n=0}^{1006}\dfrac{C_n^{1006}(2^{2013}\cot^{2n+1}2x+2014\tan^{2n+1}x)}{2n+1}+C \end{align}