## Problem Statement

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

$$
y=\arcsin x-\sqrt{1-x^{2}}, 0 \leq x \leq \frac{15}{16}
$$
## Solution

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

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

Then, using the formula above, we get:

$$
\begin{aligned}
L & =\int_{0}^{15 / 16} \sqrt{1+\left(\frac{1+x}{\sqrt{1-x^{2}}}\right)^{2}} d x= \\
& =\int_{0}^{15 / 16} \sqrt{1+\frac{1+x}{1-x}} d x= \\
& =\int_{0}^{15 / 16} \sqrt{\frac{1-x+1+x}{1-x}} d x= \\
& =\int_{0}^{15 / 16} \frac{\sqrt{2} d x}{\sqrt{1-x}}= \\
& =-\sqrt{2} \int_{0}^{15 / 16} \frac{d(1-x)}{\sqrt{1-x}}= \\
& =-\left.\sqrt{2} \cdot 2 \sqrt{1-x}\right|_{0} ^{15 / 16}=-2 \sqrt{2}\left(\sqrt{1-\frac{15}{16}}-1\right)=-2 \sqrt{2}\left(-\frac{3}{4}\right)=\frac{3}{\sqrt{2}}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\�\�\� \%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2 \%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3\%D1\%80 $\% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_17-16 \%$

Categories: Kuznetsov's Problem Book Integrals Problem 17 | Integrals

- Content is available under CC-BY-SA 3.0 .

Kuznetsov Integrals Problem 17-16 — PlusPi

Created by GeeTeatoo

![](https://cdn.mathpix.com/cropped/2024_05_22_e20a2c7468984dcb970fg-33.jpg?height=77&width=58&top_left_y=126&top_left_x=1256)

(

- 

(

(

(

(n)

(

(

$\square$

(

$\square$

(

(

$-2$

(

(

$-1$

(

(nt

![](https://cdn.mathpix.com/cropped/2024_05_22_e20a2c7468984dcb970fg-33.jpg?height=83&width=478&top_left_y=2323&top_left_x=181)

(

## Kuznetsov Integrals Problem $17-17$

## Material from PlusPi