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

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

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

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

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

Solution steps

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

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

\int [(3 - x^{2}) 3] d x

x \to 0 lim (x^{6})

x \to - \infty lim (x, 7)

\frac{\partial}{\partial x} (6 x^{2} - x + 2)

- \frac{4}{x}

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