What is the integral of 8/(sqrt(16-x^2)) ?

The integral of 8/(sqrt(16-x^2)) is 8arcsin(1/4 x)+C

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integral of 8/(sqrt(16-x^2))

\int \frac{8}{16 - x ^{2}} d x

8 arcsin (\frac{1}{4} x) + C

Solution steps

\int \frac{8}{16 - 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}{16 - 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}{4} x)

= 8 \cdot 1 \cdot arcsin (\frac{1}{4} x)

Simplify

= 8 arcsin (\frac{1}{4} x)

Add a constant to the solution

= 8 arcsin (\frac{1}{4} x) + C

y = 7 - x, x = \frac{( y - 4 ) ^{2}}{4}

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

\frac{\partial}{\partial x} (6 x + 3 y)

\frac{d}{d x} (ln (x^{2} - 3))

y = 2 x^{4} - 16 x^{2} + 9, at x = - 4

What is the integral of 8/(sqrt(16-x^2)) ?
The integral of 8/(sqrt(16-x^2)) is 8arcsin(1/4 x)+C