What is the integral from 0 to 1 of 9/(sqrt(1-x^2)) ?

The integral from 0 to 1 of 9/(sqrt(1-x^2)) is (9pi)/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral from 0 to 1 of 9/(sqrt(1-x^2))

\int_{0}^{1} \frac{9}{1 - x ^{2}} d x

\frac{9 π}{2}

Decimal Notation

14.13716 \dots

Solution steps

\int_{0}^{1} \frac{9}{1 - x ^{2}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 9 \cdot \int_{0}^{1} \frac{1}{1 - x ^{2}} d x

Use the common integral:

\int \frac{1}{1 - x ^{2}} d x = arcsin (x)

= 9 [arcsin (x)]_{0}^{1}

Compute the boundaries:

\frac{π}{2}

= 9 \cdot \frac{π}{2}

Simplify

= \frac{9 π}{2}

\frac{\partial}{\partial x} (x^{4} - 5 x y + y^{4})

f (x) = \frac{x ^{2} + 4 x + 3}{x}

y^{^{'}} tan^{- 1} (x) - y (1 + x^{2})^{- 1} = 0

\frac{d}{d x} (cos (x) + 2 sin (x) cos (x))

\int \frac{w}{1 + w ^{2}} d w

What is the integral from 0 to 1 of 9/(sqrt(1-x^2)) ?
The integral from 0 to 1 of 9/(sqrt(1-x^2)) is (9pi)/2