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

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

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

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

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

Solution steps

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

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} y)

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

Simplify

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

Add a constant to the solution

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

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