What is the integral of 3/(sqrt(16-9x^2)) ?

The integral of 3/(sqrt(16-9x^2)) is arcsin(3/4 x)+C

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integral of 3/(sqrt(16-9x^2))

\int \frac{3}{16 - 9 x ^{2}} d x

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

Solution steps

\int \frac{3}{16 - 9 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 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int \frac{1}{3} d u

= 3 \cdot \int \frac{1}{3} d u

Take the constant out:

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

= 3 \cdot \frac{1}{3} \cdot \int 1 d u

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{1}{3} \cdot 1 \cdot u

Substitute back

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

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

Simplify

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

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

Add a constant to the solution

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

y = \frac{x}{x + \frac{b}{x}}

\frac{1}{( 1 + x ) ^{3}}, 0

\frac{3}{s ^{2} + 9}

h (x) = \frac{9}{x}, at x = 2

e^{- 5 x - 3}, 5

What is the integral of 3/(sqrt(16-9x^2)) ?
The integral of 3/(sqrt(16-9x^2)) is arcsin(3/4 x)+C