Boku No Hero ep 80 Integral I just managed to catch up to an anime called Boku No hero, and to my surprise they showed an integral in ep 80-81 I think. Where the main character tried to evaluate it and got $\frac{107}{12}$ but the answer was $\frac{107}{28}$. I'm just curious how would anyone evaluate this? So the integral was 
$$I =\int^{\log(1+\sqrt{2})}_0\left(\frac{e^x-e^{-x}}{2}\right)^3\left(\frac{e^x-e^{-x}}{2}\right)^{11}dx$$
This would be the same as integrating $$I=\int^{\log(1+\sqrt{2})}_0\sinh^{14}xdx$$ But I'm not sure how to go on from here. Is there any hidden trick that I fail to see?
 A: It seems there are many typos in this adaptation of manga into anime. In Chapter 167 of manga (in both english and chinese translation), the main character and his classmates have been asked to evaluate the integral below:

(image extracted from reahheroacademia.com )
As one can see, in the second factor of the integrand, there is a '+' instead of '-' between the two exponentials. The correct integral to evaluate is
$$\mathcal{I} = \int_0^{\log(1+\sqrt{2})} \sinh^3 x\cosh^{11} x dx$$
If one zoom in above picture, one will notice the color of the blackboard is not a solid gray. It is a pattern obtained by scattering white dots over black background. It is hard to recognize the '-' in those $e^{-x}$ appear in the integrand. This may explain why in the anime, it is mistakenly to be $\hat{e}^x$.
Finally in the manga, unlike the numbers $\frac{107}{12}$ and $\frac{107}{24}$ quoted in question, the main character has evaluated the integral to $\frac{107}{14}$ and the correct answer in manga $\frac{107}{28}$ is the correct one in real world.
Let's back to the math and change variable to $u = \sinh(x)^2$. 
When $x$ changes from $0$ to $\log(1+\sqrt{2})$, $u$ increases from $0$ monotonically to $1$. Since $\cosh^2x = \sinh^2x + 1$ and $\frac{du}{dx} = 2\sinh x\cosh x$, we can evaluate the integral as:
$$\begin{align}
\mathcal{I} 
&= \frac12\int_0^{\log(1+\sqrt{2})} (\sinh^2 x)(\cosh^2 x)^5 (2\sinh x\cosh x)dx\\
&= \frac12\int_0^1 u (u+1)^5 du\\
&= \frac12\int_0^1 (u+1)^6 - (u+1)^5 du\\
&= \frac12\left[\frac{(u+1)^7}{7}-\frac{(u+1)^6}{6}\right]_0^1\\
&= \frac12\left[\frac{128-1}{7} - \frac{64-1}{6}\right]\\
&= \frac{107}{28}\end{align}$$
