What is the integral of 4/(sqrt(16-x^2)) ?

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

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integral of 4/(sqrt(16-x^2))

\int \frac{4}{16 - x ^{2}} d x

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

Solution steps

\int \frac{4}{16 - x ^{2}} d x

Take the constant out:

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

= 4 \cdot \int \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int 1 d u

= 4 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 4 \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

t^{2} y^{^{''}} + 19 t y^{^{'}} + 82 y = 0

(9, - 7), (4, - 8)

\int_{0}^{9} \frac{1}{81 - x ^{2}} d x

\frac{\partial}{\partial x} (6 x y - 5 x^{2} y)

x, 3

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