## Problem Statement

Calculate the lengths of the arcs of the curves given by equations in a rectangular coordinate system.

$$
y=\sqrt{1-x^{2}}+\arcsin x, 0 \leq x \leq \frac{7}{9}
$$
## Solution

The length of the arc of a curve defined by the equation $y=f(x) ; a \leq x \leq b$, is determined by the formula

$$
L=\int_{a}^{b} \sqrt{1+\left(f^{\prime}(x)\right)^{2}} d x
$$

Let's find the derivative of the given function:

$$
f^{\prime}(x)=\left(\sqrt{1-x^{2}}+\arcsin x\right)^{\prime}=-\frac{2 x}{2 \sqrt{1-x^{2}}}+\frac{1}{\sqrt{1-x^{2}}}=\sqrt{\frac{1-x}{1+x}}
$$

Then, using the above formula, we get:

$$
\begin{aligned}
L & =\int_{0}^{7 / 9} \sqrt{1+\left(\sqrt{\frac{1-x}{1+x}}\right)^{2}} d x= \\
& =\int_{0}^{7 / 9} \sqrt{1+\frac{1-x}{1+x}} d x= \\
& =\int_{0}^{7 / 9} \sqrt{\frac{1+x+1-x}{1+x}} d x= \\
& =\int_{0}^{7 / 9} \frac{\sqrt{2} d x}{\sqrt{1+x}}= \\
& =\left.2 \sqrt{2} \sqrt{1+x}\right|_{0} ^{7 / 9}=2 \sqrt{2}\left(\frac{4}{3}-1\right)=\frac{2 \sqrt{2}}{3}
\end{aligned}
$$

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Categories: Kuznetsov Problem Book Integrals Problem 17 | Integrals

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## Problem Kuznetsov Integrals 17-4

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