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

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

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integral of 1/(sqrt(9-(x+2)^2))

\int \frac{1}{9 - ( x + 2 ) ^{2}} d x

arcsin (\frac{1}{3} (x + 2)) + C

Solution steps

\int \frac{1}{9 - ( x + 2 ) ^{2}} d x

Apply u-substitution

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

= \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

= 1 \cdot arcsin (\frac{1}{3} (x + 2))

Simplify

= arcsin (\frac{1}{3} (x + 2))

Add a constant to the solution

= arcsin (\frac{1}{3} (x + 2)) + C

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