Just a small twist

Calculus Level 4

$\large{\int^{\frac{\pi}{4}}_{0} (\tan^{2016} (x)+\tan^{2014} (x))d(x-\frac{\lfloor x \rfloor}{1!}+\frac{\lfloor x \rfloor^{2}}{2!}-\frac{\lfloor x \rfloor^3}{3!}+\ldots)}$

If the value of the integral above equals to $\frac ab$ for coprime positive integers, find the value of $a+b$ .

The answer is 2016.

1 solution

Kunal Gupta
Jun 20, 2015

As $\dfrac{\pi}{4} <1$ ,hence $[x], [x]^2,......$ are all zero.
So, $\displaystyle \int^{\frac{\pi}{4}}_{0} (\tan^{2016} (x)+\tan^{2014} (x))d(x-\frac{[x]}{1!}+\frac{[x]^{2}}{2!}-\frac{[x]^3}{3!}+....) = \displaystyle \int^{\frac{\pi}{4}}_{0} (\tan^{2016} (x)+\tan^{2014} (x))dx$ $\displaystyle \int^{\frac{\pi}{4}}_{0} (\tan^{2014} (x)(1+ \tan^{2}(x) )dx = \displaystyle \int^{\frac{\pi}{4}}_{0} (\tan^{2014} (x)(\sec^2 (x) )dx$ $= \dfrac{\tan^{2015}{x}}{2015}$ which evaluates to $\dfrac{1}{2015}$ , so answer is: $\boxed{\boxed{ 2016}}$

It can be also derived that

$I_n + I_{n-2} = \dfrac{1}{n-1}$

$I_n = \displaystyle \int_{0}^{\frac{\pi}{4}} \tan^nx dx$

Vishwak Srinivasan - 5 years, 11 months ago

Just a small twist

The answer is 2016.

1 solution

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