What is the integral of 8/(sqrt(64-x^2)) ?

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

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integral of 8/(sqrt(64-x^2))

\int \frac{8}{64 - x ^{2}} d x

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

Solution steps

\int \frac{8}{64 - x ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{64 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 8 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 8 \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

\frac{18 x + 4 x ^{2}}{( 9 + 4 x ) ^{2}}

\int e^{- x} cos (a x) d x

\int 2 x e^{- y} d x

\frac{d}{d x} (4 cos (x) + 2 sin (x))

x \to \infty lim (π^{x})

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