Take the 2-minute tour ×  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
app