What is the integral of 1/(sqrt(8-x^2)) ?

The integral of 1/(sqrt(8-x^2)) is arcsin(1/(2sqrt(2))x)+C

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integral of 1/(sqrt(8-x^2))

\int \frac{1}{8 - x ^{2}} d x

arcsin (\frac{1}{2 2} x) + C

Solution steps

\int \frac{1}{8 - 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}{2 2} x)

= 1 \cdot arcsin (\frac{1}{2 2} x)

Simplify

= arcsin (\frac{1}{2 2} x)

Add a constant to the solution

= arcsin (\frac{1}{2 2} x) + C

x^{2} - 4 x + 6

\int \frac{1}{y ^{\frac{11}{5}}} d y

\frac{d}{d x} (\frac{x ^{2} y ^{2}}{2})

\int - 4 x^{6} - 5 x^{5} - 2 x^{2} + 5 - e^{x} d x

f (x) = arcsec (x + 2)

What is the integral of 1/(sqrt(8-x^2)) ?
The integral of 1/(sqrt(8-x^2)) is arcsin(1/(2sqrt(2))x)+C