What is the integral of (-16)/(sqrt(4-x^2)) ?

The integral of (-16)/(sqrt(4-x^2)) is -16arcsin(1/2 x)+C

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integral of (-16)/(sqrt(4-x^2))

\int \frac{- 16}{4 - x ^{2}} d x

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

Solution steps

\int \frac{- 16}{4 - x ^{2}} d x

Take the constant out:

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

= - 16 \cdot \int \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

= - 16 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= - 16 \cdot 1 \cdot u

Substitute back

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

= - 16 \cdot 1 \cdot arcsin (\frac{1}{2} x)

Simplify

= - 16 arcsin (\frac{1}{2} x)

Add a constant to the solution

= - 16 arcsin (\frac{1}{2} x) + C

What is the integral of (-16)/(sqrt(4-x^2)) ?
The integral of (-16)/(sqrt(4-x^2)) is -16arcsin(1/2 x)+C