What is the integral from 0 to 4 of 1/(sqrt(16-x^2)) ?

The integral from 0 to 4 of 1/(sqrt(16-x^2)) is pi/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 4 of 1/(sqrt(16-x^2))

\int_{0}^{4} \frac{1}{16 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{4} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{2}} 1 d u

= \int_{0}^{\frac{π}{2}} 1 d u

Integral of a constant:

\int a d x = a x

= [1 \cdot u]_{0}^{\frac{π}{2}}

Simplify

= [u]_{0}^{\frac{π}{2}}

Compute the boundaries:

\frac{π}{2}

= \frac{π}{2}

x \to 0 + lim (x^{2 x})

\frac{\partial}{\partial t} (e^{t})

n = 2 \sum \infty \frac{x ^{n}}{n}

d y + (3 x^{2} y - x^{2}) d x = 0

\int cos (x + y) d x

What is the integral from 0 to 4 of 1/(sqrt(16-x^2)) ?
The integral from 0 to 4 of 1/(sqrt(16-x^2)) is pi/2