What is area y= 3/2 ,y= 1/(sqrt(1-x^2)) ?

The answer to area y= 3/2 ,y= 1/(sqrt(1-x^2)) is sqrt(5)-2arcsin((sqrt(5))/3)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

area y= 3/2 ,y= 1/(sqrt(1-x^2))

area

y = \frac{3}{2}, y = \frac{1}{1 - x ^{2}}

5 - 2 arcsin (\frac{5}{3})

Decimal Notation

0.55393 \dots

Solution steps

Apply the area formula:

\int_{- \frac{5}{3}}^{\frac{5}{3}} \frac{3}{2} - \frac{1}{1 - x ^{2}} d x

Solve

\int_{- \frac{5}{3}}^{\frac{5}{3}} \frac{3}{2} - \frac{1}{1 - x ^{2}} d x : 5 - 2 arcsin (\frac{5}{3})

The area is:

= 5 - 2 arcsin (\frac{5}{3})

ln (\frac{e ^{2 x} ( x ^{2} + 3 )}{x + 2})

- x^{2} + 3 x - 2

\frac{d}{d x} (x e^{- b x})

\int (5 t^{2} + 3 t) d t

\frac{d}{d x} (sec (x))

What is area y= 3/2 ,y= 1/(sqrt(1-x^2)) ?
The answer to area y= 3/2 ,y= 1/(sqrt(1-x^2)) is sqrt(5)-2arcsin((sqrt(5))/3)