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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?

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    $\begingroup$ I guess they use the 'year' in such problems as a kind of a hidden hint, meaning 'obviously don't try calculating by hand, there's a trick to this' $\endgroup$ – Spine Feast Apr 27 '14 at 16:08
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    $\begingroup$ @DepeHb, sometimes substitutions are obvious to such 'tricky' questions... But I cannot see how something easy or clever or tricky can be done to solve this. Would be great to see awesome solutions to this question.. $\endgroup$ – Apurv Apr 27 '14 at 16:12
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    $\begingroup$ Is this a definite integral whose limits were omitted, or is it really an indefinite one? $\endgroup$ – apnorton Apr 27 '14 at 16:21
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    $\begingroup$ Probably about the best I could do is write it as a sum. $\endgroup$ – Mike Apr 27 '14 at 16:37
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    $\begingroup$ It can be seperable such that $$\frac{1}{2^{2014}}\int\frac{dx}{\sin^{2014}2x}-2014\int\frac{dx}{\cos^{2014}x}$$ But ı dont know what it does? $\endgroup$ – guest Apr 27 '14 at 17:10
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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}

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