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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


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up vote 2 down vote accepted

Since you know that $\cos^2x = \frac{1 + \cos(2x)}2$, try the substitution $u = x^2$.

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I have it! Thank you very much. – Zolani13 Apr 1 '12 at 5:52

Substitute $t=x^2$ , so $dt=2x dx$ , hence :

$I= \frac{1}{2} \int \limits_{0}^{\pi^2/9} \cos^2t \,dt$

Now use a fact that :

$\cos 2t=2\cos^2 t -1$

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