Determine all $x > 0$ where function $f(x) = 1 + \int_{-x}^{x}\cos(t^2) dt $ reaches local maximum. I took the derivative of this function and got $f^\prime(x) = 2\cos(x^2)$ but further i put this equal to zero but can't find out the values for which $\cos(x^2) = 0$. It would be a great help if someone tells me how to find the value where function($\cos(x^2) = 0)$?
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You have $\cos(u) = 0$ when $u$ is an odd multiple of $\frac{\pi}{2}$. – J126 Aug 16 '17 at 22:58
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Hint: have you tried looking at the unit circle? Particularly at the points where that lie on the vertical axis? – Simply Beautiful Art Aug 16 '17 at 22:59
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1@SimplyBeautifulArt yes i did but the table/graph has not all values , i mean i can see that cos(x) is zero at $\pi/2$ but it is $\cos(x^2)$. I was looking for an easy method if there is any. :-) – Afraz Salim Aug 16 '17 at 23:14
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$$\theta=x^2\\cos(\theta)=0\\theta\in\left{\dots,-\frac{3\pi}2,-\frac\pi2,\frac\pi2,\frac{3\pi}2,\dots\right}\x=\sqrt\theta$$ – Simply Beautiful Art Aug 16 '17 at 23:15