What is the integral of 6/(sqrt(1-t^2)) ?

The integral of 6/(sqrt(1-t^2)) is 6arcsin(t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 6/(sqrt(1-t^2))

\int \frac{6}{1 - t ^{2}} d t

6 arcsin (t) + C

Solution steps

\int \frac{6}{1 - t ^{2}} d t

Take the constant out:

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

= 6 \cdot \int \frac{1}{1 - t ^{2}} d t

Use the common integral:

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

= 6 arcsin (t)

Add a constant to the solution

= 6 arcsin (t) + C

What is the integral of 6/(sqrt(1-t^2)) ?
The integral of 6/(sqrt(1-t^2)) is 6arcsin(t)+C