What is the integral of (1/(sqrt(16-x^2))) ?

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

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integral of (1/(sqrt(16-x^2)))

\int (\frac{1}{16 - x ^{2}}) d x

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

Solution steps

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

Apply Trigonometric Substitution

: \int 1 d u

= \int 1 d u

Integral of a constant:

\int a d x = a x

= 1 \cdot u

Substitute back

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

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

Simplify

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

Add a constant to the solution

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

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