What is the integral of 3/(sqrt(1-9x^2)) ?

The integral of 3/(sqrt(1-9x^2)) is arcsin(3x)+C

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integral of 3/(sqrt(1-9x^2))

\int \frac{3}{1 - 9 x ^{2}} d x

arcsin (3 x) + C

Solution steps

\int \frac{3}{1 - 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}{1 - 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 (3 x)

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

Simplify

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

= arcsin (3 x)

Add a constant to the solution

= arcsin (3 x) + C

\int (x^{2} - 3 x - 4) d x

f (x) = x + 4, (12, 4)

4 y^{^{''}} + 6 y^{^{'}} - 4 y = 0

\int 1 - cos (x) d x

f (x) = \frac{x ^{3}}{6 x - 3}

What is the integral of 3/(sqrt(1-9x^2)) ?
The integral of 3/(sqrt(1-9x^2)) is arcsin(3x)+C