What is the integral of 8/(sqrt(16-(x+4)^2)) ?

The integral of 8/(sqrt(16-(x+4)^2)) is 8arcsin(1/4 x+1)+C

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integral of 8/(sqrt(16-(x+4)^2))

\int \frac{8}{16 - ( x + 4 ) ^{2}} d x

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

Solution steps

\int \frac{8}{16 - ( x + 4 ) ^{2}} d x

Take the constant out:

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

= 8 \cdot \int \frac{1}{16 - ( x + 4 ) ^{2}} d x

Apply u-substitution

: \int \frac{1}{16 - u ^{2}} d u

= 8 \cdot \int \frac{1}{16 - u ^{2}} d u

Apply Trigonometric Substitution

: \int 1 d v

= 8 \cdot \int 1 d v

Integral of a constant:

\int a d x = a x

= 8 \cdot 1 \cdot v

Substitute back

= 8 \cdot 1 \cdot arcsin (\frac{1}{4} (x + 4))

Simplify

8 \cdot 1 \cdot arcsin (\frac{1}{4} (x + 4)) : 8 arcsin (\frac{1}{4} x + 1)

= 8 arcsin (\frac{1}{4} x + 1)

Add a constant to the solution

= 8 arcsin (\frac{1}{4} x + 1) + C

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What is the integral of 8/(sqrt(16-(x+4)^2)) ?
The integral of 8/(sqrt(16-(x+4)^2)) is 8arcsin(1/4 x+1)+C