Can you generalize it?

Calculus Level 4

$\large \displaystyle \int_{0}^{8\pi} \max \left( \sin x, \sin^{-1} (\sin x) \right) \, dx= \, ?$

\pi^2-2

\frac{4 \left(\pi^2-8 \right)}{3}

\pi^2-4

\pi^2-8

1 solution

Akhil Bansal
Nov 30, 2015

Curve marked with light blue color(see image) shows the graph of given function and it repeat itself after $2\pi$ . Hence, its period is $2\pi$ .

$\large I= \displaystyle \int_{0}^{8\pi} \max \left( \sin x, \sin^{-1} (\sin x) \right) \, dx$ $\large I =4\left[ \displaystyle \int_{0}^{\pi} \sin^{-1} (\sin x) \, dx + \displaystyle \int_{\pi}^{2\pi} \sin x \, dx\right]$ $\large I = 4\left[\displaystyle \int_{0}^{\frac{\pi}{2}} x + \displaystyle\int_{\frac{\pi}{2}}^{\pi} \left(\pi - x\right) \, dx + \displaystyle\int_{\pi}^{2\pi} \sin x \, dx \right]$ Now, its elementary integration after that, $I = \pi^2 - 8$

Moderator note:

Great! Very clear explanation indicating what the various regions are.

So the general formula for integrating it from 0 to n*pi = pi^2-n

Vijay Simha - 3 years, 8 months ago

no, i guess the formula would be,

$n\cdot \dfrac{\pi^{2}}{8}-n$ , if $n$ is even, and,

$(n+1)\cdot \dfrac{\pi^{2}}{8}-(n-1)$ , if $n$ is odd.

Anirudh Sreekumar - 3 years, 8 months ago

Can you generalize it?

1 solution

Moderator note:

0 pending reports