Integrate 2

So, what I did was integration by parts. Let $u = \arccos(\sin(x^2 ) )$ and $dv = dx$ . By the formula, you get:

$x \arccos(\sin(x^2 ) ) + \int \dfrac{2x^2 \cos(x^2 )}{\sqrt{1-\sin^{2} (x^2) }}$

$= x \arccos(\sin(x^2) ) + \int \dfrac{2x^2 \cos(x^2 ) }{\sqrt{\cos^{2} (x^2)}}$

$= x \arccos(\sin(x^2) ) + \int \dfrac{2x^2 \cos(x^2 ) }{ \left| \cos (x^2)\right| }$

That last integral is tricky. You cannot cancel out the cosines due to the absolute value. The function itself is very annoying, keeps going up and down within 0 to $\pi$ . You need to then figure out where $\cos(x^2 )$ is positive and $\cos(x^2 )$ is negative because that absolute value stops you from canceling. When $\cos(x^2 )$ is positive, multiply the integral (within the correct limits of integration) by 1 and when negative, multiply integral (within correct limits of integration) by $-1$ .

So these limits of integration are:

$0, \sqrt{\dfrac{\pi}{2}}$

$\sqrt{\dfrac{\pi}{2}}, \sqrt{\dfrac{3\pi}{2}}$

$\sqrt{\dfrac{3\pi}{2}}, \sqrt{\dfrac{\pi}{2} + 2\pi}$

$\sqrt{\dfrac{\pi}{2} + 2\pi}, \pi$

In the first and third intervals, multiply the integral of $2x^2$ by positive 1. In the second and fourth intervals, multiply the integral of $2x^2$ by negative 1.

This gives us $\dfrac{\pi^{3/2}}{3} \left( \sqrt{2} - 3 \sqrt{6} + 5 \sqrt{6} - 2 \pi^{3/2}\right)$ , approximately $- 2.3378$ . http://www.wolframalpha.com/input/?i=(2%2F3)+(sqrt(pi%2F2))%5E3+-+(+(2%2F3)(sqrt(3pi%2F2))%5E3+-+(2%2F3)(sqrt(pi%2F2))%5E3++)+%2B+(++(2%2F3)(%5Csqrt(pi%2F2+%2B+2pi))%5E3+-+(2%2F3)(sqrt(3pi%2F2))%5E3)+-++((2%2F3)+pi%5E3+-+(2%2F3)(%5Csqrt(pi%2F2+%2B+2pi))%5E3)

$x \arccos(\sin(x^2) )$ from 0 to $\pi$ is about 6.3322... ) http://www.wolframalpha.com/input/?i=pi(arccos(sin(pi%5E2

Add those numbers together, and you get $\boxed{3.9945}$