What is the integral from 0 to 1 of (4-x^2)^{-3/2} ?

The integral from 0 to 1 of (4-x^2)^{-3/2} is 1/(4sqrt(3))

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integral from 0 to 1 of (4-x^2)^{-3/2}

\int_{0}^{1} (4 - x^{2})^{- \frac{3}{2}} d x

\frac{1}{4 3}

Decimal Notation

0.14433 \dots

Solution steps

\int_{0}^{1} (4 - x^{2})^{- \frac{3}{2}} d x

Apply exponent rule:

a^{- b} = \frac{1}{a ^{b}}

= \int_{0}^{1} \frac{1}{( 4 - x ^{2} ) ^{\frac{3}{2}}} d x

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{6}} \frac{1}{4 cos ^{2} ( u )} d u

= \int_{0}^{\frac{π}{6}} \frac{1}{4 cos ^{2} ( u )} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int_{0}^{\frac{π}{6}} \frac{1}{cos ^{2} ( u )} d u

Use the common integral:

\int \frac{1}{cos ^{2} ( u )} d u = tan (u)

= \frac{1}{4} [tan (u)]_{0}^{\frac{π}{6}}

Compute the boundaries:

\frac{1}{3}

= \frac{1}{4} \cdot \frac{1}{3}

Simplify

= \frac{1}{4 3}

\int_{0}^{1} (x - 1)^{2} d x

\int_{0}^{1} \frac{1}{1 + x ^{4}} d x

\int_{0}^{x^{2}} 3 sin (t) d t

\int_{1}^{5} \frac{1}{x - 3} d x

\int_{0}^{t} sin (t) d t

What is the integral from 0 to 1 of (4-x^2)^{-3/2} ?
The integral from 0 to 1 of (4-x^2)^{-3/2} is 1/(4sqrt(3))