## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\sin 1} \frac{(\arcsin x)^{2}+1}{\sqrt{1-x^{2}}} d x
$$
## Solution

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_4-15$ »

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

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## Problem Kuznetsov Integrals 4-16

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