What is the integral from 0 to 1 of 5/(sqrt(4-x^2)) ?

The integral from 0 to 1 of 5/(sqrt(4-x^2)) is (5pi)/6

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integral from 0 to 1 of 5/(sqrt(4-x^2))

\int_{0}^{1} \frac{5}{4 - x ^{2}} d x

\frac{5 π}{6}

Decimal Notation

2.61799 \dots

Solution steps

\int_{0}^{1} \frac{5}{4 - x ^{2}} d x

Take the constant out:

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

= 5 \cdot \int_{0}^{1} \frac{1}{4 - x ^{2}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} 1 d u

= 5 \cdot \int_{0}^{\frac{π}{6}} 1 d u

Integral of a constant:

\int a d x = a x

= 5 [1 \cdot u]_{0}^{\frac{π}{6}}

Simplify

= 5 [u]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{π}{6}

= 5 \cdot \frac{π}{6}

Simplify

= \frac{5 π}{6}

y = ln (x^{4} - 4095), (8, 0)

\int - sec (x) tan (x) d x

\frac{d}{d x} (ln (x + x^{2} - 9))

\frac{d}{d x} (\frac{x + 1}{x ^{3} + x - 2})

\frac{d}{d x} (x ln (5 x))

What is the integral from 0 to 1 of 5/(sqrt(4-x^2)) ?
The integral from 0 to 1 of 5/(sqrt(4-x^2)) is (5pi)/6