What is the integral of 1/(sqrt(9-64x^2)) ?

The integral of 1/(sqrt(9-64x^2)) is 1/8 arcsin(8/3 x)+C

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integral of 1/(sqrt(9-64x^2))

\int \frac{1}{9 - 64 x ^{2}} d x

\frac{1}{8} arcsin (\frac{8}{3} x) + C

Solution steps

\int \frac{1}{9 - 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 (\frac{8}{3} x)

= \frac{1}{8} \cdot 1 \cdot arcsin (\frac{8}{3} x)

Simplify

= \frac{1}{8} arcsin (\frac{8}{3} x)

Add a constant to the solution

= \frac{1}{8} arcsin (\frac{8}{3} x) + C

y (θ) = π^{4} cos (θ)

\frac{d}{d t} (9 + t)

\frac{d}{d x} (x + 2 cos (x))

x^{2} - 2 x + 8

\int x^{5} 1 + x^{3} d x

What is the integral of 1/(sqrt(9-64x^2)) ?
The integral of 1/(sqrt(9-64x^2)) is 1/8 arcsin(8/3 x)+C