What is the integral of 3/(sqrt(16-x^2)) ?

The integral of 3/(sqrt(16-x^2)) is 3arcsin(1/4 x)+C

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integral of 3/(sqrt(16-x^2))

\int \frac{3}{16 - x ^{2}} d x

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

Solution steps

\int \frac{3}{16 - x ^{2}} d x

Take the constant out:

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

= 3 \cdot \int \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

= 3 \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 3 \cdot 1 \cdot u

Substitute back

u = arcsin (\frac{1}{4} x)

= 3 \cdot 1 \cdot arcsin (\frac{1}{4} x)

Simplify

= 3 arcsin (\frac{1}{4} x)

Add a constant to the solution

= 3 arcsin (\frac{1}{4} x) + C

d er i v a t i v e x (x - 4)

\frac{d}{d x} (3 \cdot e^{\frac{x ^{3}}{9}})

\int (9 x - 2)^{- 3} d x

\int 5 r e^{\frac{r}{2}} d r

d er i v a t i v e f (x) = x^{2} ln (8 - 5 x^{2})

What is the integral of 3/(sqrt(16-x^2)) ?
The integral of 3/(sqrt(16-x^2)) is 3arcsin(1/4 x)+C