What is the integral of 1/(sqrt(26-x^2)) ?

The integral of 1/(sqrt(26-x^2)) is arcsin(1/(sqrt(26))x)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(sqrt(26-x^2))

\int \frac{1}{26 - x ^{2}} d x

arcsin (\frac{1}{26} x) + C

Solution steps

\int \frac{1}{26 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

u = arcsin (\frac{1}{26} x)

= 1 \cdot arcsin (\frac{1}{26} x)

Simplify

= arcsin (\frac{1}{26} x)

Add a constant to the solution

= arcsin (\frac{1}{26} x) + C

\frac{d}{d x} (\frac{4 x + 2}{x + 1})

\int tan^{5} (5 x) d x

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

\int sec^{2} (x) cos (x) d x

y = x^{2} - 1, y = x + 1

What is the integral of 1/(sqrt(26-x^2)) ?
The integral of 1/(sqrt(26-x^2)) is arcsin(1/(sqrt(26))x)+C