What is the integral of 9/(sqrt(4-x^2)) ?

The integral of 9/(sqrt(4-x^2)) is 9arcsin(1/2 x)+C

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integral of 9/(sqrt(4-x^2))

\int \frac{9}{4 - x ^{2}} d x

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

Solution steps

\int \frac{9}{4 - x ^{2}} d x

Take the constant out:

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

= 9 \cdot \int \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 9 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 9 \cdot 1 \cdot u

Substitute back

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

= 9 \cdot 1 \cdot arcsin (\frac{1}{2} x)

Simplify

= 9 arcsin (\frac{1}{2} x)

Add a constant to the solution

= 9 arcsin (\frac{1}{2} x) + C

x \to 0 lim (2 - x)

x^{\frac{4}{3}}

f (x) = x^{- \frac{2}{3}}

\frac{d y}{d x} = \frac{y}{2 x}, y (4) = 1

tan^{2} (2 x)

What is the integral of 9/(sqrt(4-x^2)) ?
The integral of 9/(sqrt(4-x^2)) is 9arcsin(1/2 x)+C