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

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

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

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

\frac{1}{2}

Decimal Notation

0.70710 \dots

Solution steps

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

Apply Trigonometric Substitution

: \int_{0}^{\frac{π}{4}} cos (u) d u

= \int_{0}^{\frac{π}{4}} cos (u) d u

Use the common integral:

\int cos (u) d u = sin (u)

= [sin (u)]_{0}^{\frac{π}{4}}

Compute the boundaries:

\frac{1}{2}

= \frac{1}{2}

\int_{a}^{x} f (t) d t

\int_{4}^{7} 4 d x

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

\int_{0}^{2} x^{2} + 3 d x

\int_{1}^{3} x^{2} + 2 x - 4 d x

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