The exact value of the integral $$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$ is determined to be $$\pi\ln(\sqrt{2}+1)$$. This conclusion is reached by utilizing the properties of the secant and arccosine functions, specifically that $$\sec(-x) = \sec(x)$$ and $$\arccos(-x) = \pi - \arccos(x)$$ for $$|x| < 1$$. The integral is simplified by breaking it into two symmetric parts and evaluating the integral of $$\sec(x)$$ over the interval from 0 to $$\pi/4$$.
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[sp]Since $\sec(-x) = \sec(x)$, and $\arccos(-x) = \pi - \arccos(x)$ (for $|x| < 1$), it follows that $$\int_{-\pi/4}^0 \arccos(x)\sec(x)\,dx = \int_0^{\pi/4} (\pi - \arccos(x))\sec(x)\,dx.$$ Hence $$\int_{-\pi/4}^{\pi/4}\arccos(x)\sec(x)\,dx = \pi\int_0^{\pi/4} \sec(x)\,dx = \pi\Bigl[\ln|\sec(x) + \tan(x)|\Bigr]_0^{\pi/4} = \pi\ln(\sqrt2+1).$$[/sp]eddybob123 said:Find the exact value of the following integral:
$$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$
Opalg said:[sp]Since $\sec(-x) = \sec(x)$, and $\arccos(-x) = \pi - \arccos(x)$ (for $|x| < 1$), it follows that $$\int_{-\pi/4}^0 \arccos(x)\sec(x)\,dx = \int_0^{\pi/4} (\pi - \arccos(x))\sec(x)\,dx.$$ Hence $$\int_{-\pi/4}^{\pi/4}\arccos(x)\sec(x)\,dx = \pi\int_0^{\pi/4} \sec(x)\,dx = \pi\Bigl[\ln|\sec(x) + \tan(x)|\Bigr]_0^{\pi/4} = \pi\ln(\sqrt2+1).$$[/sp]