What is the integral from 0 to 2 of 3/(sqrt(16-x^2)) ?

The integral from 0 to 2 of 3/(sqrt(16-x^2)) is pi/2

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integral from 0 to 2 of 3/(sqrt(16-x^2))

\int_{0}^{2} \frac{3}{16 - x ^{2}} d x

\frac{π}{2}

Decimal Notation

1.57079 \dots

Solution steps

\int_{0}^{2} \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_{0}^{2} \frac{1}{16 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

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

Integral of a constant:

\int a d x = a x

= 3 [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= 3 [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= 3 \cdot \frac{π}{6}

Simplify

= \frac{π}{2}

\frac{d}{d x} ((cos (x))^{\frac{π}{5}})

\int_{- 2}^{3} \frac{45}{x ^{4}} d x

\frac{d y}{d x} = \frac{x + y}{x + y + 2}

\int \frac{x + 5}{x ^{2} ( x - 2 )} d x

x \to 0 lim (ln (1))

What is the integral from 0 to 2 of 3/(sqrt(16-x^2)) ?
The integral from 0 to 2 of 3/(sqrt(16-x^2)) is pi/2