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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply Trigonometric Substitution

: \int 1 d u

= 16 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 16 \cdot 1 \cdot u

Substitute back

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

= 16 \cdot 1 \cdot arcsin (\frac{1}{4} x)

Simplify

= 16 arcsin (\frac{1}{4} x)

Add a constant to the solution

= 16 arcsin (\frac{1}{4} x) + C

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