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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

= \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 (4 x)

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

Simplify

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

Add a constant to the solution

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

d er i v a t i v e e^{c o t (4 x)}

d er i v a t i v e \frac{6 x ^{2} + 8 x + 8 x}{x}

y^{^{'}} = \frac{( y ^{2} + 2 x y )}{x ^{2}}

\int e^{3 x} sin (3 x) d x

y^{^{'}} = \frac{x}{y}, y (0) = 3

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