What is the integral from 0 to 6 of 1/(sqrt(36+t^2)) ?

The integral from 0 to 6 of 1/(sqrt(36+t^2)) is ln(1+sqrt(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 6 of 1/(sqrt(36+t^2))

\int_{0}^{6} \frac{1}{36 + t ^{2}} d t

ln (1 + 2)

Decimal Notation

0.88137 \dots

Solution steps

\int_{0}^{6} \frac{1}{36 + t ^{2}} d t

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} sec (u) d u

= \int_{0}^{\frac{π}{4}} sec (u) d u

Use the common integral:

\int sec (u) d u = ln ∣ tan (u) + sec (u) ∣

= [ln ∣ tan (u) + sec (u) ∣]_{0}^{\frac{π}{4}}

Compute the boundaries:

ln (1 + 2)

= ln (1 + 2)

\int \frac{x + 2}{x} d x

x \to - 1 lim (2 x^{2} - 4)

\frac{d}{d x} (\frac{1 + 6 x}{9 - 4 x})

y^{2} - 2 x = 7, x - y = 4

x^{5} \cdot e^{x}

What is the integral from 0 to 6 of 1/(sqrt(36+t^2)) ?
The integral from 0 to 6 of 1/(sqrt(36+t^2)) is ln(1+sqrt(2))