## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{1}{\sqrt{2}}} \frac{(\arccos x)^{3}-1}{\sqrt{1-x^{2}}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\frac{1}{\sqrt{2}}} \frac{\arccos ^{3} x-1}{\sqrt{1-x^{2}}} d x=\int_{0}^{\frac{1}{\sqrt{2}}} \frac{\arccos ^{3} x}{\sqrt{1-x^{2}}} d x+\int_{0}^{\frac{1}{\sqrt{2}}} \frac{-d x}{\sqrt{1-x^{2}}}= \\
& =\int_{0}^{\frac{1}{\sqrt{2}}}-(\arccos x)^{3} d(\arccos x)+\left.\arccos x\right|_{0} ^{\frac{1}{\sqrt{2}}}= \\
& =-\left.\frac{1}{4} \cdot(\arccos x)^{4}\right|_{0} ^{\frac{1}{\sqrt{2}}}+\arccos \frac{1}{\sqrt{2}}-\arccos 0= \\
& =-\frac{1}{4} \cdot\left(\arccos \frac{1}{\sqrt{2}}\right)^{4}+\frac{1}{4} \cdot(\arccos 0)^{4}+\frac{\pi}{4}-\frac{\pi}{2}= \\
& =-\frac{1}{4} \cdot\left(\frac{\pi}{4}\right)^{4}+\frac{1}{4} \cdot\left(\frac{\pi}{2}\right)^{4}-\frac{\pi}{4}=-\frac{\pi^{4}}{4^{5}}+\frac{\pi^{4}}{4 \cdot 2^{4}}-\frac{\pi}{4}=\frac{15 \pi^{4}}{2^{10}}-\frac{\pi}{4}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_4-25$ »

Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

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## Problem Kuznetsov Integrals 4-26

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