What is the integral of (1/(sqrt(9-x^2))) ?

The integral of (1/(sqrt(9-x^2))) is arcsin(1/3 x)+C

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integral of (1/(sqrt(9-x^2)))

\int (\frac{1}{9 - x ^{2}}) d x

arcsin (\frac{1}{3} x) + C

Solution steps

\int \frac{1}{9 - 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}{3} x)

= 1 \cdot arcsin (\frac{1}{3} x)

Simplify

= arcsin (\frac{1}{3} x)

Add a constant to the solution

= arcsin (\frac{1}{3} x) + C

f^{^{''}} (x) = (7 + 6 x)^{e^{- 5 x}}

\int x (3 - 5 x) d x

\frac{d}{d x} (cos^{3} (e^{- x}))

\int cos (\frac{mπ x}{3}) d x

\frac{\partial}{\partial y} (2 y cos (x))

What is the integral of (1/(sqrt(9-x^2))) ?
The integral of (1/(sqrt(9-x^2))) is arcsin(1/3 x)+C