Integral Question

[eddybob123]

Find the exact value of the following integral:
$$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$

 

[Opalg]

Find the exact value of the following integral:
$$\int^{\pi /4}_{-\pi /4 }\arccos(x)\sec(x)\;dx$$

Spoiler

Since $\sec(-x) = \sec(x)$, and $\arccos(-x) = \pi - \arccos(x)$ (for $|x| < 1$), it follows that \(\displaystyle \int_{-\pi/4}^0 \arccos(x)\sec(x)\,dx = \int_0^{\pi/4} (\pi - \arccos(x))\sec(x)\,dx.\) Hence \(\displaystyle \int_{-\pi/4}^{\pi/4}\arccos(x)\sec(x)\,dx = \pi\int_0^{\pi/4} \sec(x)\,dx = \pi\Bigl[\ln|\sec(x) + \tan(x)|\Bigr]_0^{\pi/4} = \pi\ln(\sqrt2+1).\)

 

[eddybob123]

Spoiler

Since $\sec(-x) = \sec(x)$, and $\arccos(-x) = \pi - \arccos(x)$ (for $|x| < 1$), it follows that \(\displaystyle \int_{-\pi/4}^0 \arccos(x)\sec(x)\,dx = \int_0^{\pi/4} (\pi - \arccos(x))\sec(x)\,dx.\) Hence \(\displaystyle \int_{-\pi/4}^{\pi/4}\arccos(x)\sec(x)\,dx = \pi\int_0^{\pi/4} \sec(x)\,dx = \pi\Bigl[\ln|\sec(x) + \tan(x)|\Bigr]_0^{\pi/4} = \pi\ln(\sqrt2+1).\)

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