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

\frac{d}{d x} (4 (x + 4)^{3})

f (x) = x^{2} - 3 x - 2, at x = 2

\frac{d}{d x} (- \frac{b}{x ^{2}})

\frac{d y}{d x} = \frac{cos ( 7 x )}{e ^{7 y}}, y (0) = 0

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

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