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

: \int 1 d u

= 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

f (x) = 4 t

x (x - 4)

n = 1 \sum \infty \frac{( - 1 ) ^{n} ( 2 ) ^{n - 1}}{( 2 ) ^{n + 1} ( n ) ^{\frac{1}{3}}}

\frac{2 x - 6}{x ^{2} + 9}

y^{^{'}} + \frac{4 y}{x} + \frac{4}{x} = 0, y (1) = - 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