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

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

Popular >

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

= \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

\int_{0}^{1} e^{5 x} - e^{x} d x

e^{c o t (4 x)}

\int \frac{( 1 + cos ^{2} ( x ) )}{cos ^{2} ( x )} d x

\frac{6 x ^{2} + 8 x + 8 x}{x}

\frac{x ^{2}}{y ^{2} - 2} y^{^{'}} = \frac{1}{2 y}, y (1) = 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