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

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

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integral of 1/(sqrt(16-4x^2))

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

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

Solution steps

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

Apply Trigonometric Substitution

= \int \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{2}{4} x)

= \frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{4} x)

Simplify

\frac{1}{2} \cdot 1 \cdot arcsin (\frac{2}{4} x) : \frac{1}{2} arcsin (\frac{x}{2})

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

Add a constant to the solution

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

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