What is the integral from 0 to 1/6 of 3/(sqrt(1-9x^2)) ?

The integral from 0 to 1/6 of 3/(sqrt(1-9x^2)) is pi/6

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integral from 0 to 1/6 of 3/(sqrt(1-9x^2))

\int_{0}^{\frac{1}{6}} \frac{3}{1 - 9 x ^{2}} d x

\frac{π}{6}

Decimal Notation

0.52359 \dots

Solution steps

\int_{0}^{\frac{1}{6}} \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_{0}^{\frac{1}{6}} \frac{1}{1 - 9 x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} \frac{1}{3} d u

= 3 \cdot \int_{0}^{\frac{π}{6}} \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_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= 3 \cdot \frac{1}{3} [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

3 \cdot \frac{1}{3} [1 \cdot u]_{0}^{\frac{π}{6}} : [u]_{0}^{\frac{π}{6}}

= [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= \frac{π}{6}

\frac{d x}{d t} = x t

y^{^{''}} - 2 y^{^{'}} + 1 y = 0, y (0) = 8, y^{^{'}} (0) = 2

\frac{\partial}{\partial y} (- 2 e^{x} cos (yz))

\int (\frac{1}{2 x}) d x

\frac{d}{d x} (F (x^{4}))

What is the integral from 0 to 1/6 of 3/(sqrt(1-9x^2)) ?
The integral from 0 to 1/6 of 3/(sqrt(1-9x^2)) is pi/6