What is the integral of 1/(sqrt(1-64x^2)) ?

The integral of 1/(sqrt(1-64x^2)) is 1/8 arcsin(8x)+C

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integral of 1/(sqrt(1-64x^2))

\int \frac{1}{1 - 64 x ^{2}} d x

\frac{1}{8} arcsin (8 x) + C

Solution steps

\int \frac{1}{1 - 64 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{8} d u

= \int \frac{1}{8} d u

Take the constant out:

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

= \frac{1}{8} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= \frac{1}{8} \cdot 1 \cdot u

Substitute back

u = arcsin (8 x)

= \frac{1}{8} \cdot 1 \cdot arcsin (8 x)

Simplify

= \frac{1}{8} arcsin (8 x)

Add a constant to the solution

= \frac{1}{8} arcsin (8 x) + C

\int \frac{2 x + 3}{x ^{3} + x ^{2} - 2 x} d x

n = 2 \sum \infty \frac{1}{n} (- 1)^{n}

y^{2} = 16 - x, (y + 2)^{2} = x + 4

\frac{e ^{s}}{7 s + 1}

f (x) = (x e^{2 x - 4}), x = 2

What is the integral of 1/(sqrt(1-64x^2)) ?
The integral of 1/(sqrt(1-64x^2)) is 1/8 arcsin(8x)+C