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

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

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

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

\frac{1}{4} arcsin (2 x) + C

Solution steps

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

Apply Trigonometric Substitution

: \int \frac{1}{4} d u

= \int \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{4} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

y = e, y = e^{x}, y = e^{- x}

y^{^{''}} + 6 y^{^{'}} + 25 y = 0, y (0) = 2, y^{^{'}} (0) = 14

\int sin (X) d X

x \to 2 + lim (\frac{- 1}{x ^{2} - 4})

\frac{\partial}{\partial x} (x^{2} + y^{2} + 2)

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