# An integral for the New Year 2016

I have built this integral with the purpose of presenting a question. I find interesting and pleasant to readers MSE for the New Year 2016 and expecting to see different methods of solution.

I can confirm that this has been the case by the welcome that has been given and the two motivated answers it have had.

Calculate:

$$\large \int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-x} }{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x-2016}}\mathrm dx$$

• After a standard change of variables, the answer seems to be 2016. – Hmm. Dec 31 '15 at 18:46
• @SoumyaSinhaBabu: Go ahead dear friend. – Piquito Dec 31 '15 at 18:48
• Observation - $\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a+b-x)dx$. – Hmm. Dec 31 '15 at 18:51
• Is this a challenge, where you know the answer? Please provide more context. – Rory Daulton Dec 31 '15 at 20:57
• It was in fact a challenge, a pleasant and pertinent one I feel. Tomorrow I´ll try to edit with my deficient English (I built the question myself for the New Year). – Piquito Jan 1 '16 at 1:56

## 2 Answers

Let, $$I=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-x}}{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x- 2016}}\ dx\tag 1$$ Now, using the property of definite integral: $\int_a^bf(x)\ dx=\int_{a}^bf(a+b-x)\ dx$, one should get

\begin{align*} I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot 2016-(4\cdot2016 -x)}}{\sqrt[5]{3\cdot 2016-(4\cdot2016 -x)}+\sqrt[5]{(4\cdot2016 -x)- 2016}}\ dx\\[3ex] I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{x-2016}}{\sqrt[5]{x-2016}+\sqrt[5]{3\cdot2016 -x}}\ dx\\[3ex] I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\ dx\tag 2\\[6ex] \end{align*}

Now, adding (1) & (2), one should get

\begin{align*} I+I&=\int_{2016}^{3\cdot 2016}\left(\frac{\sqrt[5]{3\cdot 2016-x}}{\sqrt[5]{3\cdot 2016-x}+\sqrt[5]{x- 2016}}+\frac{\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\right)\ dx\\[3ex] 2I&=\int_{2016}^{3\cdot 2016}\frac{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}{\sqrt[5]{3\cdot2016 -x}+\sqrt[5]{x-2016}}\ dx\\[3ex] I&=\frac12\int_{2016}^{3\cdot 2016}\ dx\\[3ex] &=\frac12(3\cdot 2016-2016)\\[3ex] &=\color{red}{2016} \end{align*}

Define $$I:=\int_{k}^{3k}\frac{\left(3k-x\right)^{1/5}dx}{\left(3k-x\right)^{1/5}+\left(x-k\right)^{1/5}},\,k:=2016.$$ The substitution $x=k\left(1+2\sin^{2}t\right)$ gives \begin{align} I &= \int_{0}^{\pi/2}\frac{4k\sin t\cos^{7/5}t\,dt}{\sin^{2/5}t+\cos^{2/5}t}\\[3ex] &= \int_{0}^{\pi/4}\frac{4k\sin t\cos t\,dt}{1+\tan^{2/5}t}+\int_{\pi/4}^{\pi/2}\frac{4k\sin^{3/5}t\cos^{7/5}t\,dt}{1+\cot^{2/5}t}. \end{align}

The replacement $t\mapsto \frac{\pi}{2}-t$ in the second integral gives \begin{align} I&= \int_{0}^{\pi/4}\frac{4k\sin t\cos t\,dt}{1+\tan^{2/5}t}+\int_{0}^{\pi/4}\frac{4k\sin^{7/5}t\cos^{3/5}t\,dt}{1+\tan^{2/5}t}\\[3ex] &= \int_{0}^{\pi/4}4k\sin t\cos t\,dt.\\[2ex] \textrm{Thus, }\qquad\qquad\\[2ex] I &= \int_{0}^{\pi/4}2k\sin2t\,dt\\[3ex] &= \left[-k\cos2t\right]_{0}^{\pi/4}\\[3ex] &=k. \end{align}

• A more complete answer than the one posted above, as you have generailsed to a limit $k$. nice =1 – Kevin Oct 18 '16 at 10:16