MHB Exact Value of Integral: $\arccos^2({\pi \over 4})$

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The integral $$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$ can be evaluated using symmetry properties of the functions involved. It is established that $$\int_{-\pi/4}^{\pi/4}\arccos(x)\sec(x)\,dx$$ equals $$\pi\int_0^{\pi/4} \sec(x)\,dx$$ due to the relationship between $\sec(-x)$ and $\sec(x)$, and $\arccos(-x)$ and $\arccos(x)$. The final result of the integral is $$\pi\ln(\sqrt2+1)$$ after evaluating the integral of the secant function. This demonstrates an effective method for calculating definite integrals involving inverse trigonometric functions. The discussion highlights the importance of symmetry in simplifying complex integrals.
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Find the exact value of the following integral:
$$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$
 
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eddybob123 said:
Find the exact value of the following integral:
$$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$
[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]
 
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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]

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