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

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

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

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

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

Solution steps

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

16 - (2 x + 2)^{2} = 2 - x^{2} - 2 x + 3

= \int \frac{1}{2 - x ^{2} - 2 x + 3} d x

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{- x ^{2} - 2 x + 3} d x

Complete the square

- x^{2} - 2 x + 3 : - (x + 1)^{2} + 4

= \frac{1}{2} \cdot \int \frac{1}{- ( x + 1 ) ^{2} + 4} d x

Apply u-substitution

= \frac{1}{2} \cdot \int \frac{1}{- u ^{2} + 4} d u

Apply Trigonometric Substitution

= \frac{1}{2} \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= \frac{1}{2} \cdot 1 \cdot v

Substitute back

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

Simplify

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

Add a constant to the solution

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

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