What is the integral of 1/(sqrt(9-u^2)) ?

The integral of 1/(sqrt(9-u^2)) is arcsin(1/3 u)+C

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integral of 1/(sqrt(9-u^2))

\int \frac{1}{9 - u ^{2}} d u

arcsin (\frac{1}{3} u) + C

Solution steps

\int \frac{1}{9 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= \int 1 d v

Integral of a constant:

\int a d x = a x

= 1 \cdot v

Substitute back

v = arcsin (\frac{1}{3} u)

= 1 \cdot arcsin (\frac{1}{3} u)

Simplify

= arcsin (\frac{1}{3} u)

Add a constant to the solution

= arcsin (\frac{1}{3} u) + C

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What is the integral of 1/(sqrt(9-u^2)) ?
The integral of 1/(sqrt(9-u^2)) is arcsin(1/3 u)+C