Consider the integral
$\int_0^{\pi/3}x\cos^2(x^2)dx $
Here, I see two terms: x, and $\cos^2(x^2)$. I would normally use integration by parts, but I run into major issues because of the $\cos^2(x^2)$, specifically the $x^2$.
I take $\cos^2(x^2)$ as $u$, x as $dv$, and after the simplifying of some constants, I have
$\frac12 x^2\cos^2(x^2) + 2\int_0^{\pi/3}x^3\cos(x^2)\sin(x^2)$
*because I found the derivative of $\cos^2(x^2)$ was $-4x\cos(x^2)\sin(x^2)$ and the integral of x as $\frac{x^2}2$
but you can see, this new integral isn't very helpful. I can't use parts with an integral with three terms being multiplied, so I'm kind of stuck.
I do know that $\cos^2x = \frac{1 + \cos(2x)}2$. That could be useful, but I don't know if this applies to different forms of $x$ such as $x^2$, like it does for $ax$, as $a$ as a constant.
How can I find the integral of these kinds of functions, where trigonometric functions have squared or cubed composed functions inside? Any help would be appreciated.
Much thanks
-Zolani